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The functional derivative of the action isFor the Hamiltonian functionand a time-independent Lagrangian,Therefore

The gravitational three-body Lagrangian is

Put $M_{12}=m_1+m_2$ and $M=M_{12}+m_3$. Solving the definitions of the Jacobi coordinates for three particles givesSubstitution into the kinetic energy makes all cross terms cancel:where

The pair separations areHence the gravitational potential energy iswhich is independent of $\mathbf{c}$. Thus $\mathbf{c}$ is an ignorable coordinate andThe center of mass moves uniformly because the isolated system has no external force.

Finally, substituting the inverse coordinate transformation intoagain cancels every cross term and yieldsThis is the angular momentum decomposition in three-body Jacobi coordinates.

Taking downward distance as positive, the Lagrangian is

At equilibrium the lower spring supports only the lower mass, while the upper spring supports both masses:Hence

Let $q_i=z_i-z_i^{(0)}$. The linear terms cancel by the equilibrium equations, and the quadratic Lagrangian iswith the small-oscillation mass and stiffness matrices

The generalized eigenvalue problem for small oscillations isIts characteristic equation issoCorresponding displacement eigenvectors are

They are not orthogonal in the ordinary Euclidean inner product. They are orthogonal in the kinetic-energy, or mass-matrix, inner product:This is mass-matrix orthogonality of normal modes.

For a general Lagrangian,Along an Euler-Lagrange trajectory,so energy is conserved when $L$ has no explicit time dependence.

For the given Lagrangian, write $v=|\dot{x}|$. At points where $v\ne 0$,The Euler-Lagrange equation is thereforeIn expanded form its left-hand side isThe momentum has fixed magnitude,so this is constant without needing to solve the equation of motion. The Hamiltonian is

The singular Legendre transform of a degree-one velocity Lagrangian applies here: $p$ determines only the direction of $\dot{x}$, not its magnitude, and the velocity Hessian is not invertible. Thus the usual inverse Legendre transform cannot reconstruct the Lagrangian from this Hamiltonian.

The objective is strictly convex, so the minimizer is unique. The Slater condition makes the Karush-Kuhn-Tucker conditions necessary and sufficient. Absorb the box constraints into the Euclidean projection onto a convex set and attach a scalar multiplier $\nu$ to $a^{T}x=b$. Stationarity over the box is equivalent towhile primal feasibility requires $a^T{x}^{\ast }=b$. Coordinatewise, these conditions areThey are also sufficient because they minimize the Lagrangian over the box and satisfy the equality constraint. Thus the projection onto a box-constrained hyperplane reduces to solving the displayed one-dimensional continuous, nonincreasing equation for $\nu$. The multiplier need not be unique on a flat interval, but the projected vector is unique.

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).

Let $\theta$ be the angle from the downward vertical to the hoop's centre, measured at the suspension point, and let $\varphi$ be the absolute angle from the downward vertical to the bead's radius from the hoop centre. The bead position is the sum of two vectors of length $a$. Including the hoop's moment of inertia about the pivot giveswhileThus one suitable Lagrangian is

Let $q=(q_1,q_2,q_3)$ and $n=(1,1,1)$. Sincethe Lagrangian depends on $q$ only through $|q{|}^2$ and $(q\cdot n{)}^2$, and on the velocity through $|\dot{q}{|}^2$. Rotations about the $n$-axis preserve all three quantities, proving the invariance.

The axially symmetric isotropic-mass Lagrangian therefore conserves the axial angular momentumup to an irrelevant normalization of $n$. Since $L$ has no explicit time dependence, conservation of energy from time-translation invariance gives the second conserved quantity

Write $f(r)=1+r^2/a^2$. The geodesic Lagrangian isIts $\theta$ equation isSpherical symmetry lets us rotate the conserved angular-momentum plane into $\theta =\pi /2$. Equivalently, initial data $\theta =\pi /2$, $\dot{\theta }=0$ solve this equation for all $\tau$.

The cyclic coordinates giveUsing the timelike normalization $g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }=-1$ givesAfter separating the constant terms,

The Lagrangian isWriting $T=2m+M$, the center of mass isAll internal forces cancel, so $\ddot{Y}=-g$ andIf the molecule is released from rest, the middle term vanishes.

Fermat’s principle makes $\int ds/c$ stationary. Writing the ray as $x=x(y)$ gives a Lagrangian independent of $y$, and its first integral is $\sin \theta /c=K$. At the origin $K=(1/\sqrt{2})/(1/\sqrt{3})=\sqrt{3/2}$. At $x=1$, $c=1/\sqrt{2}$, so $\sin \theta =\sqrt{3}/2$ and the forward angle is $\theta =\pi /3$.

A Lagrangian is regular when its velocity Hessian matrix is invertible. The inverse function theorem then makes the fiber derivative $p_i=L_{v^i}$ locally invertible, allowing a local Legendre transform in mechanics to a Hamiltonian. If that fiber derivative is globally invertible, the Lagrangian is called a hyperregular Lagrangian. Regularity alone does not assert global invertibility, and singular Lagrangians may give constrained rather than ordinary unconstrained Hamilton's equations.