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For a variation $y+\epsilon \eta$ with $\eta (0)=\eta (1)=0$, stationarity givesThe boundary term vanishes. The fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equation $d{F}_{y^{\prime }}/dx-F_y=0$.

Differentiating the proposed first integral givesThus it is constant along an extremal when $F_x=0$, proving the Beltrami identity.

For the positive branch relevant to the endpoint data, $F=\sqrt{1+y^{\prime 2}}/y$. The identity becomesThe Euler-Lagrange equation itself gives $y{y}^{\prime \prime }=-(1+y^{\prime 2})$, so $(y^2{)}^{\prime \prime }=-2$. Integrating, or solving the first integral, yields a circle $y^2+(x-c{)}^2=R^2$. The equal endpoint values force $c=1/2$ and $R^2=5/4$. Hence the unique positive smooth extremal isIt stays positive on the entire interval and satisfies the second-order equation, including its turning point. Merely differentiating the first integral after dividing by $y^{\prime }$ would need extra care there. Geometrically this is an arc of a hyperbolic geodesic in the upper half-plane model.

Write $I[y]={\int }_a^{b}G(x,y,y^{\prime })dx$. For endpoint-zero variations $\eta$, the first variations areFor a normal constrained extremal, choose a variation ${\eta }_0$ with $\delta I({\eta }_0)\ne 0$. Given any $\eta$, use the implicit function theorem to choose a small correction in the ${\eta }_0$ direction so that $y+t\eta +s(t){\eta }_0$ preserves the constraint. Differentiating it and the stationarity condition at zero givesIntegrate the derivative-of-variation term by parts. Since the variations vanish at both endpoints, the boundary term is zero. The fundamental lemma of the calculus of variations then givesThe regularity condition on the constraint is what permits the coefficient of $F$ to be normalized to 1. In a degenerate case, the general multiplier statement uses ${\lambda }_{0}F-{\lambda }_{1}G$ with the two coefficients not both zero; the source's displayed form is the normal case.

For a string described as a function graph $r(x)=(x,y(x))$, its arc-length element is $|r^{\prime }(x)|dx=\sqrt{1+y^{\prime }(x{)}^2}dx$. Inextensibility therefore givesThis also forces $l\ge 2a$, with equality only for the straight function graph.

Write $K[y]={\int }_a^{b}G(x,y,y^{\prime })dx$. At a regular constrained extremal, the first variation of $K$ is not identically zero. Choose a variation $\zeta$ for which $\delta K[\zeta ]\ne 0$; the implicit function theorem supplies a correction in the $\zeta$ direction to each other variation so that $K$ remains one. Stationarity of $J$ on these corrected variations implies $\delta J=\lambda \delta K$ for one scalar $\lambda$.

Put $L=F-\lambda G$. Integration by parts gives, for an arbitrary twice continuously differentiable variation $\eta$,Taking variations with zero endpoint values and using the stated fundamental lemma of the calculus of variations gives the Euler-Lagrange equation. The remaining endpoint values are free and independent, giving the natural boundary conditions:If the constraint is not regular, the general multiplier statement permits an abnormal pair ${\lambda }_{0}F-{\lambda }_{1}G$ with ${\lambda }_0=0$; the source's general normal-form derivation implicitly excludes that case. In its quadratic example, regularity follows automatically from normalization: $\delta K[y]=2\int w{y}^2=2$.

For that example $L=p{y}^{\prime 2}-q{y}^2-\lambda w{y}^2$, so the equation and endpoint conditions reduce toMultiply the equation by $y$ and integrate. The boundary term $[p{y}^{\prime }y{]}_a^b$ vanishes, givingThe last integral equals one, so $\lambda =J[y]$, the functional value at the constrained extremal.

Finally multiply the equation for $y_1$ by $y_2$ and the equation for $y_2$ by $y_1$, then subtract and integrate. The derivative terms give only the vanishing boundary term $[p(y_2{y}_1^{\prime }-y_1{y}_2^{\prime }){]}_a^b$, while the $q$ terms cancel. Therefore $({\lambda }_1-{\lambda }_2){\int }_a^{b}w{y}_1{y}_{2}dx=0$. For distinct multipliers this proves

At a regular constrained extremal, the first variations satisfy $\delta I=\lambda \delta J$ for a constant Lagrange multiplier $\lambda$. One way to justify this is to choose a variation for which $\delta J\ne 0$, use it to correct any other variation to preserve the constraint by the implicit function theorem, and differentiate $I$ along that corrected family. This gives the same ratio $\lambda$ for all variations. The exceptional case $\delta J=0$ for every variation requires the usual separate abnormal-extremal analysis.

Set $L=F-\lambda G$. Integration by parts in its first variation givesTaking compactly supported variations and applying the fundamental lemma of the calculus of variations yields the Euler-Lagrange equation. For the quadratic integrands, it isor $\mathcal{L}y=\lambda ry$ with $\mathcal{L}y=-(p{y}^{\prime }{)}^{\prime }-qy$. If endpoint values are free, independent endpoint variations give the natural conditionsAt a prescribed endpoint the variation instead vanishes and no natural condition is added there.

Under these natural conditions, integration by parts givesHence $\boxed{I=\lambda J}$. For two solutions satisfying the same natural conditions, subtract their cross-multiplied equations and integrate:The final equality is the vanishing endpoint boundary form. Thus distinct eigenvalues giveThe same argument works for other common homogeneous boundary conditions making that boundary form zero.

In the Legendre case, $((1-x^2)y_n^{\prime }{)}^{\prime }+{\lambda }_n{y}_n=0$. Comparing coefficients of $x^n$ in a degree-$n$ solution gives ${\lambda }_n=n(n+1)$, so different degrees have different eigenvalues. A degree-zero solution is $y_0=C_0$. A linear solution $Ax+B$ must be orthogonal to the nonzero constant solution, so $B=0$ and $y_1=C_{1}x$. Write $y_2=A{x}^2+Bx+C$. Orthogonality to $x$ gives $B=0$, and orthogonality to $1$ gives $2A/3+2C=0$. ThereforeThe constants are free normalization factors; for solutions of the stated exact degrees they are nonzero. The natural endpoint conditions hold automatically for these polynomials because $p(\pm 1)=0$.

Take a smooth variation $u+\epsilon v$ with $v=0$ on the boundary. Differentiating under the integral gives the first variationIntegration by parts removes the derivatives of $v$, and the boundary contribution vanishes. The fundamental lemma of the calculus of variations therefore givesFor a regular constraint $J[u]=c$, the Lagrange multiplier rule says that an extremal satisfies $\delta I=\lambda \delta J$ for a constant $\lambda$. Equivalently, apply the preceding Euler–Lagrange equation to $F-\lambda G$. Regularity means that $\delta J$ is not identically zero as a linear functional on the admissible variations. For the quadratic constraint below this holds at every nonzero admissible $u$; a zero constraint leaves only $u=0$.

In polar coordinates the constrained integrand isIts Euler–Lagrange equation is $-2\lambda ru-{\partial }_r(2r{u}_r)-{\partial }_{\theta }(2u_{\theta }/r)=0$, orThe field is single-valued and periodic in $\theta$. Substitution $u=f(r)\cos (n\theta )$ with $n\in \{0,1,2,\dots \}$ givesThe centre is part of the disk, so one must also select the regular radial solution: for $n\ge 1$, smoothness gives $f(r)=O(r^n)$ and the regular derivative behaviour; for $n=0$, $f$ is bounded and $f^{\prime }(0)=0$. A singular solution at zero does not define an admissible smooth field on the disk.

For fixed $n$, let $c_n={\int }_0^{2\pi }{\cos }^2(n\theta )d\theta$, so $c_0=2\pi$ and $c_n=\pi$ for $n\ge 1$. The angular derivative term has integral $n^2{c}_n$, including zero when $n=0$. ConsequentlyThe one-dimensional Euler–Lagrange equation for $\hat{I}-\lambda \hat{J}$ isexactly the same radial equation. The common factor $c_n$ cancels.

For two regular solutions with the same $n$ and different multipliers, multiply the equation for $f_1$ by $f_2$, multiply that for $f_2$ by $f_1$, subtract and integrate. The result isAt $a$, both functions vanish; at zero, regularity makes the boundary term tend to zero. Since the multipliers are distinct,This is weighted Sturm–Liouville orthogonality with weight $r$.

Finally, multiply one radial equation by $f$ and integrate by parts. The same endpoint conditions yieldMultiplication by $c_n$ provesFor any nonzero admissible solution the energy on the left before subtraction is positive, so its multiplier is the positive Rayleigh quotient $\lambda =\hat{I}/\hat{J}$.

The first weak derivative $D_{j}u=v$ is characterized by ${\int }_{\Omega }u{\partial }_j\varphi =-{\int }_{\Omega }v\varphi$ for every $\varphi \in C_c^{\infty }(\Omega )$. The fundamental lemma of the calculus of variations in the locally integrable form says that a function pairing to zero with every such test is zero almost everywhere. Applied to the difference of two candidate derivatives, it proves uniqueness. The Sobolev space consists of $u\in L^p$ with all first weak derivatives in $L^p$, with normFor the one-dimensional tent, integration by parts on each side of one has cancelling junction terms because the function is continuous. Thus $u^{\prime }=1$ on $(0,1)$ and $u^{\prime }=-1$ on $(1,2)$, almost everywhere. Both it and its derivative are bounded, so it belongs to every stated Sobolev space; its displayed norm to the power $p$ is $2/(p+1)+2$.

For $u=x_1/r$ in the punctured ball, ordinary differentiation givesThe function is bounded and $|D_{j}u|\le 2/r$, so polar integration gives finite derivative $L^p$ norms when ${\int }_0^1{r}^{n-1-p}dr<\infty$, namely $p<n$. To verify these are weak derivatives across the origin, integrate by parts outside a ball of radius $\epsilon$ against a compactly supported smooth test. The inner boundary term is at most a fixed constant times ${\epsilon }^{n-1}$, since $u$ is bounded. It tends to zero for $n>1$. The volume terms converge by local integrability of the displayed derivative. Thus $u\in W^{1,p}(B)$ for $1\le p<n$ with exactly these derivatives. Its value at the origin can be chosen arbitrarily.

The first variation isThe endpoint terms vanish. The fundamental lemma of the calculus of variations gives $xu=0$ almost everywhere, hence $u=0$ on $(0,1)$. A Lipschitz representative is then identically zero and cannot have terminal value one. Therefore there are no admissible solutions of the stationarity equation.

Let $\eta$ be a smooth variation with $\eta ={\eta }^{\prime }=0$ at both endpoints, and consider $y_{ϵ}=y+ϵ\eta$. stationary point gives the first variationIntegrate the middle term once by parts and the last term twice. All boundary terms vanish because both $\eta$ and ${\eta }^{\prime }$ vanish there. Assuming the derivatives needed for the classical formula, the given fundamental lemma of the calculus of variations yieldsWithout assuming those classical derivatives initially, the same calculation with compactly supported variations states this equation weakly; sufficient regularity then gives the classical equation.

For the specific integrand, $F_y=-2y$, $F_{y^{\prime }}=0$, $F_{y^{\prime \prime }}=2y^{\prime \prime }$, so the equation is $y^{(4)}-y=0$. Its weak solutions here are smooth, and its general classical solution isPut $\ell =\pi /2$, $c=\cosh \ell$, $s=\sinh \ell$. The conditions at zero give $C=-A$ and $D=1-B$. Those at $\ell$ giveUsing $c^2-s^2=1$, solve to obtain $A=c/2$, $B=(1-s)/2$, $C=-c/2$, $D=(1+s)/2$. Thus an extremal isSubstitution checks all four boundary values. The additive term $x^2$ in the integrand contributes no variation in $y$, so it does not alter this equation.

Take an admissible variation $u+ϵ\eta$ with $\eta =0$ on the boundary. Differentiating the functional at an extremum givesIntegration by parts removes the derivatives from $\eta$; the boundary term vanishes. The fundamental lemma of the calculus of variations yieldsFor the specified squared-gradient energy this is $2\Delta u=0$, so $u$ is harmonic. The affine function $\ell =ax+by+cz$ is harmonic and satisfies the boundary data.

To verify a global minimum, write $u=\ell +v$ with zero boundary trace. ThenThe middle term vanishes by integration by parts, since $\Delta \ell =0$ and $v=0$ at the boundary. Hence $I[u]\ge I[\ell ]$, andThis proves minimality directly, rather than relying only on a necessary Euler-Lagrange equation.

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.

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 $y$ be a smooth constrained minimizer. For any smooth variation $h$ vanishing at the endpoints, use the normalized admissible curve $y_{\epsilon }=(y+\epsilon h)/\sqrt{J[y+\epsilon h]}$. Since $J[y]=1$, differentiating at zero givesIn particular this holds for every compactly supported test function $h$ in $(0,1)$. Integration by parts and the fundamental lemma of the calculus of variations givewhich is the required Bessel differential equation with $\boxed{\lambda =I[y]}$. The lemma used here says that a continuous function whose integral against every compactly supported smooth test function is zero must itself vanish. No multiplier-existence theorem is needed: the explicit normalization produces the variation directly.

The endpoint $0$ causes no divergent energy for these variations: smoothness and $y(0)=h(0)=0$ imply $y,h=O(x)$. Also $J[v]>0$ for every nonzero admissible $v$, and $v/\sqrt{J[v]}$ has unit $J$ and energy $I[v]/J[v]$. Thus minimizing $I$ with $J=1$ is equivalent to minimizing the scale-invariant Rayleigh quotient $I/J$. The zero function is excluded whenever this quotient is written.

Hamilton's principle states that the physical path makes the action $S={\int }_{t_i}^{t_f}L(\mathbf{q},\dot{\mathbf{q}},t)dt$ stationary under differentiable variations $\delta q_i$ vanishing at the fixed endpoints. Its conventional “least action” name does not guarantee a minimum. Expanding to first order and integrating the velocity term by parts givesThe endpoint term vanishes. By the fundamental lemma of the calculus of variations, the coefficients of the independent interior variations must vanish, yieldingFor example, the zero path of a harmonic oscillator with endpoints zero can have both positive and negative second variations when the elapsed time exceeds a half-period; stationary action then need not be a least value. The Euler-Lagrange condition is the local statement that the variational principle actually supplies.

Hamilton's principle makes the action $S[q]={\int }_{t_1}^{t_2}L(q,\dot{q},t)dt$ stationary under variations vanishing at the endpoints. It need not be a strict minimum. Varying the action and integrating the velocity variation by parts givesThe endpoint term vanishes. Since the interior variations are arbitrary, the fundamental lemma of the calculus of variations yields

If the fields are continuously differentiable and obey the pointwise conservation law, integration by parts against a compactly supported test function gives the weak identity, with no boundary terms. Conversely, integrate the weak identity by parts for a continuously differentiable field:for every test function. The fundamental lemma of the calculus of variations, applied to each component, makes the continuous integrand vanish at every point. Hence the classical and weak formulations are equivalent for continuously differentiable fields.

Let $P^{\mu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }h)$ and $Q^{\mu \nu }=\partial \mathcal{L}/\partial ({\partial }_{\mu }{\partial }_{\nu }h)$. Define the second coefficient with ordered repeated indices and $Q^{\mu \nu }=Q^{\nu \mu }$, so thatThis convention avoids double-counting a mixed derivative such as $h_{,12}=h_{,21}$. The first derivative term needs one integration by parts. The second derivative term needs two:Combining the terms givesFor compactly supported variations, or fixed $\delta h$ and its first derivatives at the boundary, the last integral vanishes. The fundamental lemma of the calculus of variations then yields the Euler-Lagrange field equation with second derivatives:The plus sign of the final term comes from the two integrations by parts. In general this gives fourth-order equations, although the order in a particular coordinate depends on which second derivatives occur in the density.

Vary the action $S=\int dtdx\mathcal{L}$ with compactly supported $\delta \varphi$, or impose vanishing $\delta \varphi$ and its first derivatives on the spacetime boundary. The first variation isOne integration by parts on the first-derivative term and two on the second-derivative term remove all derivatives from $\delta \varphi$. The boundary terms vanish, leavingThe fundamental lemma of the calculus of variations gives the Euler-Lagrange field equation with second derivatives:Repeated indices run over time and space. The coefficient of the symmetric second derivative is defined by the displayed variation; if ${\varphi }_{tx}$ is instead treated as a single independent mixed derivative variable, it is counted once rather than twice.