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Treat $\varphi$ and $\bar{\varphi }$ as independent fields. The Lagrangian density isThe Euler-Lagrange equation for $\bar{\varphi }$ isorThe equation from variation of $\varphi$ is its complex conjugate.

Put $q=\delta \theta$, $p=\delta \varphi$, so $\theta =\pi /2-q$ and $\varphi =p$. The quadratic Lagrangian is obtained by takingThe two real fluctuations form a coordinate and its canonical momentum rather than two independent modes. Their Euler-Lagrange equations combine to giveand similarly for $p$. The ferromagnetic magnon therefore has the quadratic dispersion relation

The area element is $\sqrt{\det I^S}dxdy=Wdxdy$, soFor the Lagrangian $L(p,q)=\sqrt{1+p^2+q^2}$, stationarity under every compactly supported variation gives the Euler-Lagrange equationMultiplication by $W^3$ expands this intorecovering part (d).

For the p-energythe first variation in the direction $\psi \in C_c^{\infty }(\Omega )$ isAn integration by parts therefore gives the Euler-Lagrange equationwhich is the p-Laplacian equation. In the notation of the question one takes $a^{ij}(x,z,\eta )=|\eta {|}^{p-2}{\delta }^{ij}$.

For $\eta \ne 0$, the principal coefficient matrix isIts eigenvalue in directions orthogonal to $\eta$ is $|\eta {|}^{p-2}$, while its eigenvalue parallel to $\eta$ is $(p-1)|\eta {|}^{p-2}$. The coefficients are $C^{1,\alpha }$ away from $\eta =0$, and the condition number there is at most $p-1$. On every region where $0<m\le |Du|\le M$, this gives uniform ellipticity with constants depending on $m$, $M$, and $p$. At $Du=0$ all principal eigenvalues vanish, so the operator is degenerate there and is not strictly elliptic on a domain containing a critical point.

The Euler-Lagrange equation isTime-translation invariance gives the conserved energyIndeed, after integrating the spatial term by parts and imposing finite-energy boundary conditions,

Take $z$ upward along the straight rod. The part above height $z$ has weight $\lambda g(L-z)$, so, with tensile force positive, the internal tension is compressive:For a small transverse displacement $X(z)$, the quadratic bending and gravitational energies areThe Euler-Lagrange equation for this functional iswhich is exactly

Clamping at the bottom fixes displacement and slope:At the free upper end, bending moment and transverse force vanish:Since $\sigma (L)=0$, the four boundary conditions are

Set $u=X^{\prime }$. Integrating the field equation once and using the free-end shear condition givesor, with $s=L-z$,Introduce the dimensionless similarity coordinateand write $u={\eta }^{1/3}F(\eta )$. Direct substitution reduces the equation toThis is the Bessel differential equation of order $1/3$, so

The free-moment condition is $u^{\prime }(L)=0$. As $\eta \to 0$,The second term has nonzero limiting $z$ derivative, so the free-end condition forces $b=0$. The free-shear condition $u^{\prime \prime }(L)=0$ then follows from the differential equation. At the clamp, $u(0)=0$, givingLet $j_{-1/3,1}$ be the smallest positive zero of this Bessel function. The first self-buckling threshold isor equivalently

For $y\in \partial \Omega$, let $b_y^{-}\le g\le b_y^{+}$ be the affine barriers from the bounded slope condition. Their constant gradients satisfy the Euler-Lagrange equation, so they minimize the autonomous convex functional for their own boundary values. The comparison principle givesSince both barriers equal $u(y)$ at $y$ and are $K$-Lipschitz,for $x\in \Omega$ and $y\in \partial \Omega$. The supplied boundary-to-interior criterion now gives $\mathrm{Lip}(u)\le K$.

Part c supplies a bounded slope condition constant $K$. Choose $k>K$. Parts a and d give a constrained minimizer with $\mathrm{Lip}(u)\le K<k$, and part b makes it a minimizer over all of ${\mathrm{Lip}}_g(B_1)$.

For $F(\xi )=\sqrt{1+|\xi {|}^2}$, the Euler-Lagrange equation is the minimal surface equation for a graphThe Lipschitz bound confines $Du$ to a compact set on which $D^{2}F$ is uniformly positive definite, so the equation is uniformly elliptic. Interior regularity gives $C^{1,\alpha }$ first, and repeated Schauder estimates then give smoothness in the interior.

For an arbitrary smooth variation $u_t=u+t\phi$, differentiation under the integral givesSince $M$ is compact without boundary, integration by parts turns this intoThe fundamental lemma of the calculus of variations therefore gives the Euler-Lagrange equationor equivalently $-{\Delta }_{g}u=f$ for the Laplace-Beltrami operator.

The p-Laplacian is the nonlinear divergence-form operatorIt is the Euler-Lagrange operator of the p-energy. For $p>2$ it is degenerately elliptic where $Du=0$.

Using the positive polar molecular field $h_i=\delta F/\delta p_i$ and velocity gradient ${\partial }_i{v}_j$, the reversible order-parameter stress consists of a distortion contribution and rotational and alignment contributions:For a local free-energy density, a representative is ${\Sigma }_{ij}^{(1)}=(f-p_k{h}_k){\delta }_{ij}-f_{{\partial }_i{p}_k}{\partial }_j{p}_k$. The Euler-Lagrange equation verifies its divergence. Integration by parts gives $\delta F=\int {\Sigma }_{ij}^p{\partial }_i{u}_j$ for an incompressible displacement without surface work; the corresponding mechanical power has the opposite sign. Pressure contributions may be reassigned in an incompressible flow.

For real $f\in L^2({\mathbb{R}}^3)$, this energy functional on $H^1({\mathbb{R}}^3)$ is coercive and attains its minimum by the direct method in the calculus of variations. The nonnegative potential term is weakly lower semicontinuous by local Sobolev compactness gives lower semicontinuity of a nonnegative integral. Its Euler-Lagrange equation is $-\Delta u+u+\sin u=f$ in the weak solution sense. The extra linear restoring term screens the Sine-Gordon equation nonlinearity. The scalar potential $V(s)=s^2/2+1-\cos s$ is convex because $V^{\prime \prime }(s)=1+\cos s\ge 0$; hence any weak critical point is a minimizer. The elliptic regularity estimate for $-\Delta +1$, together with $|\sin u|\le |u|$, gives $u\in H^2$. The Sobolev inequality then puts $u$ in $W^{1,6}$, so Morrey's inequality and uniformly continuous integrable functions vanish at infinity give a continuous representative tending to zero.

For a bounded operator on a real Hilbert space, $⟨Lv,v⟩=⟨Sv,v⟩$ with $S=(L+L^{\ast })/2$. Consequently the Euler-Lagrange equation of $⟨Lv,v⟩-2⟨f,v⟩$ is $Sv=f$. It is $Lv=f$ only when $L$ is self-adjoint or when the relevant solution also annihilates the skew part. For example $L=\left(\begin{array}{cc}1& -1\\ 1& 1\end{array}\right)$ has positive quadratic form $\Vert v{\Vert }^2$, but that form contains no information about its skew part.

For a Lorentzian metric and timelike $v=dx/du$, put $L=\sqrt{-g(v,v)}>0$ and $\mathcal{L}=L^2/2$. The Euler-Lagrange equations of $\mathcal{L}$ give ${\nabla }_{v}v=0$, while those of $L$ give ${\nabla }_{v}v=(\dot{L}/L)v$. Taking $ds/du=L$ removes the latter tangential acceleration and gives a unit timelike tangent. Thus the two actions have the same geodesic images; the quadratic action singles out an affine parameter. They do not have the same solutions for an arbitrary fixed parameter: $t=u^2$, $\mathbf{x}=0$, $u>0$, in Minkowski spacetime solves the length equation but not the energy equation. Null curves are excluded since $L=0$.

For the leading periodic propulsive force of a periodic planar filament and power of a periodic planar filament functionals, the Euler-Lagrange equation for $F_x+\Gamma (\dot{W}-{\dot{W}}_0)$ gives $({\xi }_{\perp }-{\xi }_{\parallel })y_{xt}+\Gamma {\xi }_{\perp }{y}_{tt}=0$. For nonzero thrust, $y_t$ obeys an advection equation and the active travelling wave is $f(x-ct)$, with $c=({\xi }_{\perp }-{\xi }_{\parallel })/(\Gamma {\xi }_{\perp })$. A static periodic shape may be added without changing leading force or power. Stationarity is a necessary extremum condition, not by itself an existence theorem for a global maximum.