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For a variation $y+\epsilon \eta$ preserving both endpoint values and slopes, $\eta ={\eta }^{\prime }=0$ at the endpoints. The first variation isIntegrate the last term twice and the middle term once. All boundary terms vanish, givingThe fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equationFor the present functional this becomes $1+y-y^{(4)}=0$. Its general solution is $y=-1+A\cosh x+B\sinh x+C\cos x+D\sin x$. The left endpoint gives $C=-A$ and $D=-B$. The right endpoint then givesThe coefficient determinant is $2(1+\cosh \pi )>0$, so $A=1$, $B=0$. The stationary function is $y_{\ast }(x)=-1+\cosh x-\cos x$.

For any admissible perturbation $\eta$, exact quadratic expansion, followed by the stationarity equation and integration by parts, givesApply the supplied Poincare inequality first to $\eta$, then to ${\eta }^{\prime }$, whose endpoint values also vanish:Thus $J[y_{\ast }+\eta ]\le J[y_{\ast }]$ for every admissible perturbation, proving a global maximum through the clamped second-variation inequality.

Let $Q$ be the configuration smooth manifold, and let $L(t,q,v)$ be a smooth Lagrangian on its tangent bundle. For a path with fixed endpoints, its action isThe principle of stationary action requires the first variation to vanish for every fixed-endpoint variation. In a coordinate chart take $q_s=q+s\eta$, with $\eta (a)=\eta (b)=0$. Differentiation under the integral and integration by parts giveRepeated coordinate indices are summed. The endpoint term vanishes. Since compactly supported variations can be chosen independently in each coordinate, the fundamental lemma of the calculus of variations yields the Euler-Lagrange equationsConversely these equations make the displayed first variation zero for every fixed-endpoint variation. Variations localized in coordinate charts establish the same assertion for paths on $Q$. Thus the Euler-Lagrange equations are exactly the stationary-path condition, not necessarily a condition for an action minimum.

The principle of stationary action says that the physical path makes the action stationary under smooth variations $q+\eta h$ with fixed endpoints $h(t_1)=h(t_2)=0$. It need not minimize the action. Differentiating and integrating by parts givesThe fundamental lemma of the calculus of variations then gives the Euler-Lagrange equation

Vary $y$ by $y+\epsilon \eta$, where the clamped endpoint conditions require $\eta ={\eta }^{\prime }=0$ at both endpoints. The first variation of the quadratic elastic energy isBoth integrations by parts have zero boundary terms. Since the variation is arbitrary in the interior, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation. Varying $z$ independently gives the second equation:The clamped endpoint values accompany these equations. If $A$ or $B$ depends on $x$, it must remain inside the derivatives.

Fix a reference path ${\gamma }_{\ast }$ from $L_0$ to $L_1$. By path connectedness, a path $\gamma$ can be joined to it by a smooth homotopy strip $u:[0,1{]}_s\times [0,1{]}_t\to M$ withSmooth homotopies may be used by the relative smoothing theorem; piecewise smooth homotopies give the same integrals. Orient the square by $ds\wedge dt$ and define the Lagrangian path-area functional byHere $\mathcal{A}$ denotes the functional called $A$ in the question. The minus sign is the convention for which downward gradient flow is $J$-holomorphic on the standard oriented strip.

For relative symplectic-area independence, glue two choices, with one orientation reversed, along their common reference and final paths. The resulting oriented surface is a relative $2$-cycle $Z$ in $(M,L_0\cup L_1)$. Since $H_2(M,L_0\cup L_1)=0$, it bounds a relative $3$-chain: as an absolute chain, $Z=\partial B+C$ for a chain $C$ supported in $L_0\cup L_1$. The symplectic form is closed, and it restricts to zero on each Lagrangian submanifold. Using piecewise smooth relative chains, Stokes theorem givesEquivalently, integration of the closed symplectic form defines the zero pairing on the zero relative homology group. The two strips therefore define the same value of $\mathcal{A}$.

Choosing a different reference path changes every value by the negative area of one fixed connecting strip, independent of $\gamma$. Changing the initially assigned reference value also adds a constant. Path connectedness ensures that this is one global constant, rather than an independent constant on each component.

For the first variation, let $\xi (t)$ be a variation vector field along $\gamma$, with $\xi (0)\in T_{\gamma (0)}{L}_0$ and $\xi (1)\in T_{\gamma (1)}{L}_1$. Differentiate the defining strip integral. Stokes theorem, or the variation formula for the integral of a closed differential form, reduces it to the final edge; the side edges give zero because their variations and tangent vectors lie in the Lagrangian submanifolds. With the chosen sign,If this vanishes for every admissible $\xi$, it in particular vanishes for every field supported in $(0,1)$. Nondegeneracy of the symplectic form, together with the fundamental lemma of the calculus of variations, forces $\dot{\gamma }(t)=0$ in the interior and hence everywhere by smoothness. Conversely, a constant path makes the displayed integral zero. The constant must lie in both endpoint submanifolds. ThusNo transverse-intersection assumption is required; the critical set can be empty or have positive dimension.

A compatible almost complex structure is a smooth bundle map $J:TM\to TM$ such thatThese conditions makea Riemannian metric. Its symmetry follows from the $J$-invariance and skew symmetry of $\omega$, and its positivity is the final compatibility condition. It also satisfies $g(Jv,w)=\omega (v,w)$.

Use the formal $L^2$ metric on the path space,The variation formula becomesThus a formal downward flow $s\mapsto u(s,\cdot )$ satisfiesPut the standard complex structure $j$ on the strip, with $j{\partial }_s={\partial }_t$. The J-holomorphic curve equation $du\circ j=J\circ du$ says $u_t=J{u}_s$, which is equivalent to the displayed flow equation because $J^2=-I$. The converse is identical: a smooth J-holomorphic curve on the strip with the specified Lagrangian boundary conditions gives a formal downward trajectory of $\mathcal{A}$.

As a sign check, along such a strip,The drop in $\mathcal{A}$ therefore equals the positive symplectic area swept out by the strip. The word formal matters: the $L^2$ calculation gives the interior equation and its Lagrangian boundary conditions; it does not assert that arbitrary smooth initial paths produce a well-posed ordinary flow on the smooth path space. In particular, the expression $J\dot{\gamma }$ need not satisfy the endpoint tangent restrictions for an arbitrary initial path. The claimed correspondence concerns smooth solutions of the strip equation.

Take a smooth variation $u_{\epsilon }=u+\epsilon \varphi$ with $\varphi =0$ on the boundary. Assume enough differentiability for the following operations, for example a smooth bounded open set, $u\in C^2$, and a twice continuously differentiable integrand. Differentiation under the integral givesBy integration by parts, the boundary term vanishes and the first variation isStationarity for every such variation, including every compactly supported smooth one, and the fundamental lemma of the calculus of variations give the Euler-Lagrange equationFor the quadratic integrand, $L_u=2k^{2}u$, $L_{u_x}=2u_x$, and $L_{u_y}=2u_y$. Thuswith the prescribed Dirichlet boundary condition. For real $k$, a stationary function is also the unique minimizer, if it exists: for any nonzero admissible $\varphi$, the linear term vanishes and $I[u+\varphi ]-I[u]={\int }_{\mathcal{D}}(|\nabla \varphi {|}^2+k^2{\varphi }^2)>0$. This last positivity uses the zero boundary values; it also holds at $k=0$.

For variations vanishing at both endpoints, differentiating the action and using integration by parts givesThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationsFor the specified Lagrangian, the gauge transformation changes it byFor arbitrary $F$, it is a total derivative precisely when $\boxed{\alpha =-1}$. The corresponding change in action is then an endpoint term whose variation is zero. In general time-dependent notation, the equations areThe electric field $-\nabla \varphi -{\partial }_t\mathbf{A}$ and magnetic field $\nabla \times \mathbf{A}$ are unchanged by the gauge transformation with $\alpha =-1$. Although the original potentials have no time dependence, their transformed versions generally do; allowing this is essential to the stated invariance.

For a variation $y+\epsilon \eta$ with $\eta ={\eta }^{\prime }=0$ at both endpoints, the first variation isTwo applications of integration by parts remove derivatives from $\eta$. The boundary terms vanish, and the fundamental lemma of the calculus of variations gives the higher-order Euler-Lagrange equation

For the stated integrand,so the equation reduces toThe general solution of this linear ordinary differential equation isDecay and finiteness of the integral force $A=B=0$; $y(0)=1$ gives $C=1$, and $y^{\prime }(0)=2$ gives $D=3$. Hence the unique stationary function is

The Dirichlet boundary condition is built into the first component's space, while the Neumann boundary condition is natural. Thus a weak solution is a pairsuch that for every $(v,z)\in H_0^1(U)\times H^1(U)$,If $u,w$ are $C^2$ up to the boundary, taking compactly supported test functions and applying the fundamental lemma of the calculus of variations gives both differential equations pointwise in $U$. Membership of $H_0^1(U)$ gives $u=0$ on $\partial U$. Applying integration by parts to the second identity and using its differential equation leavesfor every smooth boundary trace $z$. Hence $\partial w/\partial \nu =0$ on $\partial U$, so the equations and both boundary conditions hold classically.

Introduce a Lagrange multiplier $\lambda$ for the normalization and varyFor a smooth variation $u+\epsilon \eta$ with $\eta =0$ on $\partial D$, integration by parts givesThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equation

Multiplying by $u$ and integrating, while usingthe divergence theorem and $u=0$ on the boundary giveThe normalization is one, so the multiplier equals the stationary value:

Take a variation $x+\epsilon \xi$ with $\xi (a)=\xi (b)=0$. Expanding the action givesAn integration by parts and the fixed endpoint conditions give the first variationThe fundamental lemma of the calculus of variations therefore yields the Euler-Lagrange equationand the second variation is

Linearizing the equation of motion about $x$ givesso the Jacobi equation isWhen $u$ has no zero on $[a,b]$,The total derivative integrates to zero because $\xi (a)=\xi (b)=0$, hence

For the simple harmonic oscillator, the Jacobi equation is $\ddot{u}+{\omega }^{2}u=0$. Put $t_0=(a+b)/2$ and chooseIf $b-a<\pi /\omega$, then $|\omega (t-t_0)|<\pi /2$ throughout $[a,b]$, so $u$ is positive there. The preceding square identity proves that the classical path is a local minimum of the action whenever the elapsed time is less than half an oscillation period.

Let $y_{\epsilon }=y+\epsilon \eta$, where the variation $\eta$ satisfies $\eta (a)=\eta (b)=0$. Differentiating the functional at $\epsilon =0$ gives its first variationUsing integration by parts and the fixed endpoint conditions,If this vanishes for every admissible $\eta$, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation

If $\varphi \in C^2(\overline{U}{)}^2$, take $v\in C_c^{\infty }(U{)}^2$ and integrate the gradient term by parts. The weak identity becomesThe fundamental lemma of the calculus of variations gives $-\Delta \varphi +A\varphi =F$ pointwise. Membership in $H_0^1(U{)}^2$, together with continuity up to the boundary, says that the boundary trace is zero, so the Dirichlet condition also holds classically.

Take a variation $y+\epsilon \eta$, where the differentiable function $\eta$ obeysbecause both $y$ and its first derivative have fixed endpoint values. The first variation of the functional isApplying integration by parts once to the second term and twice to the third givesThe endpoint conditions on $\eta$ and ${\eta }^{\prime }$ make every boundary term zero. Since the remaining integral vanishes for every admissible variation, the fundamental lemma of the calculus of variations yields the higher-order Euler-Lagrange equation

Let $u\mapsto u+\epsilon \eta$ and $v\mapsto v+\epsilon \xi$, where the variations vanish on the boundary. The first variation isApplying integration by parts to the four derivative terms and discarding the boundary contributions givesThe variations $\eta$ and $\xi$ are independent and arbitrary in the interior. The fundamental lemma of the calculus of variations therefore gives the two Euler-Lagrange equations for two fields