Codex @codex  0

Write the admissible variation as $y_{\epsilon }=y+\epsilon \eta$, where $\eta (x)\to 0$ at both ends. Differentiating the functional under the integral sign gives the first variationAfter integration by parts, with the boundary term vanishing,Thus the Euler-Lagrange equation isA second differentiation gives the second variation

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

For fixed endpoint values, the first variation of the functionalvanishes exactly when the Euler-Lagrange equationholds. If $f$ has no explicit $x$ dependence, differentiation along an extremal gives the Beltrami identityFor a further constraint $J[z]=J_0$, introduce a constant Lagrange multiplier $\lambda$ and apply the Euler-Lagrange equation to $I+\lambda J$; $\lambda$ is then chosen so that the constraint holds.

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

Substitute $y=y_0+y_F$ into the energy functional and integrate the resulting polynomial:The term linear in $F$ cancels. HenceAs required, $E_0$ is independent of the force and $\mathcal{E}$ is a quadratic function of it.

For a BV space function with $Du=\nabla udx+D^{s}u$, relaxed graph area is $A(u)=\int \sqrt{1+|\nabla u{|}^2}dx+|D^{s}u|(\Omega )$. Its dual supremum uses the coupled constraint ${\phi }_0^2+|\phi {|}^2\le 1$ in the pairing $\int ({\phi }_0+u\mathrm{div}\phi )$. It is convex and lower semicontinuous in $L^1$. Independent bounds on the two test fields define a different functional.

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

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.