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In polar coordinates,With the change of radial coordinate $u=\log r$, the detection functional becomesIt is therefore $\lambda$ times ordinary path length in the $(u,\theta )$-plane. The shortest path is the line segment from $(\log A,0)$ to $(\log B,\alpha )$, soReturning to $r$ gives the logarithmic spiralIts minimum detection probability isThe functional depends only on the geometric path and not its parametrization, so tiptoeing and running give the same probability.

For the improved sensor, omit the irrelevant positive factor $\lambda$ and use the LagrangianThe coordinate $\theta$ is cyclic, so its conjugate momentum is conserved. Equivalently,is constant. Because $L$ has no explicit time dependence, the associated conservation of energy gives the constantMultiplying by $r$ and using $h=r\dot{\theta }$ gives the required equation

Let each mass be $m$ and each spring constant be $k$, with the $x_i$ denoting longitudinal displacements from equilibrium. The Lagrangian isThe inverse coordinate transformation isSubstitution givesThus the Euler-Lagrange equations separate into a symmetric pair,and an antisymmetric pair,

The symmetric normal modes are:

For antisymmetric modes, the squared dimensionless frequencies are the eigenvalues ofnamely $2\mp \sqrt{2}$. They giveandIn the original coordinates their mode shapes are respectivelyand

The coordinates $\varphi$ and $\psi$ are cyclic, so their canonical momenta are conserved:Because the Lagrangian has no explicit time dependence, the conserved energy isThese are the three integrals of motion requested.

If $\mathbf{r}$ is measured in the rotating frame, its inertial velocity is $\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. The masses producing the gravitational field are stationary in this frame, so their potential energy is a time-independent function $V(\mathbf{r})$. Thus kinetic minus potential energy gives the stated LagrangianPut $\mathbf{v}=\dot{\mathbf{r}}+\mathbf{\omega }\times \mathbf{r}$. ThenSubstitution in the Euler-Lagrange equation yieldsThe extra terms are the Coriolis acceleration, Euler acceleration, and centrifugal acceleration of a rotating reference frame.

The canonical momentum and Hamiltonian areAccordingly, Hamilton's equations are

For a regular Lagrangian, define the canonical momentum $p_i=\partial L/\partial {\dot{q}}_i$ and invert this relation for $\dot{q}(q,p,t)$. Its Legendre transform is the HamiltonianSince $Ldt=p\cdot dq-Hdt$, the action isFor independent variations with $\delta q(t_1)=\delta q(t_2)=0$,Stationarity for arbitrary $\delta p$ and $\delta q$ gives Hamilton's equations $\dot{q}={\partial }_{p}H$ and $\dot{p}=-{\partial }_{q}H$.

The kinetic energy of the two equal masses isThe extensions of the left, coupling, and right springs are respectively $x_1$, $x_2-x_1$, and $x_2$. Their potential energy is thereforeThus the mechanical Lagrangian $L=T-U$ has the requested quadratic form with the symmetric small-oscillation mass and stiffness matrices

For the equality constraint, the Lagrangian and dual function areand the Lagrangian dual problem is ${\sup }_{z\in {\mathbb{R}}^m}q(z)$. Weak duality says $q(z)\le p^{\ast }$ for every $z$. Strong duality means the dual supremum equals the primal infimum, usually with a dual maximizer. A sufficient convex constraint qualification is that $f$ be proper, closed and convex and that some $x\in \mathrm{ri}(\mathrm{dom}f)$ satisfy $Ax=b$.

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 $\theta =\pi /3$.

For small oscillations with quadratic kinetic and potential energies, a normal mode has a fixed spatial amplitude pattern and one angular normal frequency. Its amplitude is a normal coordinate; choosing all modes diagonalizes the two quadratic forms and decouples their equations. A zero frequency describes a free symmetry motion.

Let $q_j=R[{\theta }_j-j\pi /2]$ be the small signed arc displacements, with cyclic indices $j=0,1,2,3$. The springs follow arcs, so their extensions are $q_{j+1}-q_j$. The Lagrangian of the normal modes of a four-bead spring ring isThe stiffness matrix divided by $k$ is the cycle graph Laplacian, with diagonal entries two, adjacent entries minus one, and zero opposite entries. An orthonormal basis of mode vectors and their frequencies isThe normal mode coordinates are $Q_{\alpha }=u_{\alpha }\cdot q$, with equations ${\ddot{Q}}_{\alpha }+{\omega }_{\alpha }^2{Q}_{\alpha }=0$. The zero mode rotates all beads together without stretching a spring and has $Q_0=A+Bt$. In each of the two degenerate middle modes, opposite beads have opposite signed arc displacements while the other pair remains stationary. Any fixed linear combination is another mode of that frequency. The last mode alternates the signed displacement of adjacent beads. The radius cancels from the frequencies because arc length supplies the same factor $R^2$ to the angular kinetic and spring energies.

Take a variation $y_{\epsilon }=y+\epsilon \eta$ with $\eta (a)=\eta (b)=0$. Differentiating the functional at zero and using integration by parts givesThe boundary term vanishes. Since the remaining integral must vanish for every smooth endpoint-zero variation, the fundamental lemma of the calculus of variations gives the Euler-Lagrange equation $d(f_{y^{\prime }})/dx=f_y$.

For the mechanical problem put $A(r)=1-2m/r>0$. The Lagrangian is independent of $\theta$, a cyclic coordinate, so its conjugate momentum $\partial L/\partial \dot{\theta }=2r^2\dot{\theta }=2J$ is conserved. Therefore $J$ is constant, including the factor of two between $J$ and the canonical momentum for this normalization of $L$.

For the radial coordinate, $\partial L/\partial \dot{r}=2\dot{r}/A$ and $\partial L/\partial r=-{\dot{r}}^2{A}^{\prime }/A^2+2r{\dot{\theta }}^2-\mu A^{\prime }/A^2$. Its Euler-Lagrange equation isFor a circular orbit $r=R>2m$, the radial velocity and acceleration vanish. Using $A^{\prime }(R)=2m/R^2$ and $\dot{\theta }=J/R^2$ leaves $2J^2/R^3=2m\mu /[R^{2}A(R{)}^2]$, henceThis is the circular orbit condition for a radial kinetic Lagrangian applied to the coefficients here.