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In proper-time gauge the particle Lagrangian is $L=m(-{\dot{t}}^2+a^2{\dot{\mathbf{x}}}^2-1)/2$, now with dots referring to proper time $s$. The canonical spatial momentum isThe FRW metric is independent of all spatial coordinates. Hence $\partial L/\partial x^i=0$ and the Euler-Lagrange equation gives $d{p}_i^C/ds=0$. This conserved quantity is the comoving momentum.

The mass-shell mass shell constraint gives $(dt/ds{)}^2-a^2|d\mathbf{x}/ds{|}^2=1$. For future-directed motion and proper peculiar velocity $\mathbf{v}=ad\mathbf{x}/dt$, it follows that $dt/ds=(1-v^2{)}^{-1/2}=\gamma$. ThereforeThe kinetic momentum here is the physical momentum in an FRW universe, measured in the orthonormal frame of a comoving observer. It redshifts as $a^{-1}$ irrespective of whether the particle is relativistic or nonrelativistic.

In the equatorial plane, the Killing symmetries give conserved quantities $E=(1-2M/r)\dot{t}$ and $L=r^2\dot{\varphi }$. The null normalization givesFor $L\ne 0$, put $u=1/r$; then $du/d\varphi =-\dot{r}/L$. ConsequentlyDifferentiating gives $u^{\prime \prime }+u=3M{u}^2$, initially where $u^{\prime }\ne 0$ and then at turning points by continuity. The circular solution follows directly from the radial equation as well. The ratio $|L/E|$ is the impact parameter of the asymptotic null orbit.

In Lagrangian mechanics, let a one-parameter group on $Q$ have generator ${\xi }_Q={\xi }^i(q){\partial }_{q^i}$. Its tangent lift acts on $TQ$ asIf $L$ is invariant, ${\xi }_{TQ}L=0$. Along an Euler–Lagrange trajectory, the Noether conserved quantity for a mechanical point symmetry isbecauseThus the conserved quantity is explicitly derived, not merely attached to the name of a symmetry. If the change of $L$ is the total derivative $dB/dt$, then $p_i{\xi }^i-B$ is conserved. For a general point transformation with time component $\tau$, the conserved expression is $p_i{\xi }^i-E_L\tau -B$, provided the variation of $Ldt$ is $dB$. Time-translation invariance of an autonomous Lagrangian therefore yields conservation of energy.

In Hamiltonian mechanics, a cotangent-lifted action preserves $\theta$ and hence $\omega$. For its infinitesimal generator ${\xi }_{T^{\ast }Q}$, Cartan's magic formula giveswhere $J_{\xi }=p_i{\xi }^i$. These components define the canonical moment map $J:T^{\ast }Q\to {\mathfrak{g}}^{\ast }$, with $⟨J,\xi ⟩=J_{\xi }$. If $H$ is invariant, $\{H,J_{\xi }\}=0$, so ${\dot{J}}_{\xi }=\{J_{\xi },H\}=0$. This is the Hamiltonian form of Noether's theorem, and the Legendre transform in mechanics identifies its charges with the Lagrangian ones.

More generally, a Hamiltonian group action is an action whose infinitesimal symplectic generators have globally defined Hamiltonian functions. The existence of those functions is an assumption, not a consequence of every symplectic action. For example, translation on a symplectic two-torus preserves $dq\wedge dp$ but contraction with ${\partial }_q$ gives $dp$, which is closed and not globally exact. Such a symmetry need not have a globally single-valued moment-map component. On $T^{\ast }Q$, the cotangent lift does have the canonical components just derived. A Hamiltonian symmetry preserving $H$ gives a conserved generator; the global momentum-map hypotheses must be stated.

With the stated time rescaling, $\dot{x}={\epsilon }^3{u}^{\prime }$ and $\ddot{x}={\epsilon }^4{u}^{\prime \prime }$, where primes refer to $\tilde{t}$. Dividing by ${\epsilon }^4$ givesAt $\epsilon =0$, put $v=u^{\prime }$ and define the HamiltonianIts orbital derivative is $H^{\prime }=v(u^2-\kappa )+(\kappa -u^2)v=0$, so this is a conserved quantity. The Hamiltonian system has $u^{\prime }=H_v=v$, $v^{\prime }=-H_u=u^2-\kappa$.

For $\kappa >0$, set $s=\sqrt{\kappa }$. The potential $U(u)=\kappa u-u^3/3$ has a minimum at $-s$ and a maximum at $s$. The corresponding center equilibrium and saddle equilibrium have energies $H_c=-2s^3/3$ and $H_s=2s^3/3$. Levels $H_c<H<H_s$ contain closed ovals around $(-s,0)$, as well as separate unbounded components on the right. At $H=H_s$, the center's ovals limit to a homoclinic orbit to the saddle, turning at $u=-2s$. Its equation isThe same energy also has external unbounded saddle separatrices for $u>s$. Levels above $H_s$ pass over the potential barrier and are open. Along the upper halves $u^{\prime }=v>0$, so the closed loops are traversed clockwise. If $\kappa =0$, the two critical points coalesce into a degenerate equilibrium; if $\kappa <0$, the potential is strictly decreasing and there is no center or homoclinic loop.

For small $\epsilon$, the conserved quantity from part (b) instead obeys $H^{\prime }=-\epsilon (\lambda +u)v^2$. The Melnikov energy-balance method evaluates the first-order net change along the unperturbed homoclinic orbit:Using the homoclinic curve $v^2=\frac{2}{3}(s-u{)}^2(u+2s)$, these are integrals of $vdu$ on its two halves. Their weighted mean is $I_1/I_0=-5s/7$. Thus the first-order splitting vanishes at $\lambda =5s/7$. Since its derivative with respect to $\lambda$ is $-I_0\ne 0$, the implicit function theorem gives a nearby curve where the stable and unstable saddle separatrices reconnect. This is the homoclinic balance for a quadratic-force oscillator and therefore a global, rather than local, bifurcation.

Returning to the original parameters, $q={\epsilon }^2\lambda$ and $\sqrt{p}={\epsilon }^2\sqrt{\kappa }$. With fixed positive $\kappa$ and $\epsilon \to 0$, the homoclinic bifurcation curve has leading formThe energy integral locates the leading curve; the perturbed orbit itself is no longer a constant-$H$ curve.

The product of the magnetic field with the velocity in this question is the three-dimensional cross product. Fix ${ϵ}_{123}=1$ and write $K^i{}_j={ϵ}^i{}_{kj}{B}^k$, so $K\mathbf{v}=\mathbf{B}\times \mathbf{v}$. In coordinates $y^0=t$, $y^i=x^i$ on ${\mathbb{R}}^3\times \mathbb{R}$, prescribe an affine connection by the following Christoffel symbols:with all other Christoffel symbols zero. These are smooth globally in the given Cartesian coordinates. The lower-index symmetry makes this a torsion-free connection. It is the geometrization of the Lorentz force by an affine connection; no condition on the curl of $\mathbf{E}$ or the divergence of $\mathbf{B}$ is required for this construction.

An affinely parametrized geodesic, with primes denoting differentiation with respect to $\lambda$, satisfiesOn the branch $t^{\prime }=c\ne 0$, use $t$ itself as an affine parameter. Dividing the spatial equation by $c^2$ givesConversely, each solution of this equation makes $\lambda =t$, $y(t)=(t,\mathbf{x}(t))$ an affinely parametrized geodesic of the constructed affine connection. Therefore its image is also an unparametrized geodesic. A general change of parameter adds a term proportional to the tangent in the geodesic equation, leaving the curve unchanged. The branch $t^{\prime }=0$ is not a trajectory with time as parameter.

For the metric realization, assume $\mathbf{E}=-\nabla U$ and $\nabla \cdot \mathbf{B}=0$. Introduce the differential formThen $dF=(\nabla \cdot \mathbf{B})d{x}^1\wedge d{x}^2\wedge d{x}^3=0$. The global Poincaré lemma on the contractible space ${\mathbb{R}}^3$ supplies a one-form $A=A_{i}d{x}^i$ with $dA=F$, equivalently a magnetic vector potential with $\nabla \times \mathbf{A}=\mathbf{B}$. An explicit radial-gauge potential for a divergence-free magnetic field isThis is the radial homotopy formula for a closed two-form and is smooth even at the origin. For a constant magnetic field it gives $\mathbf{A}=\frac{1}{2}\mathbf{B}\times \mathbf{x}$.

Consider the Eisenhart-Duval lift with the Lorentzian metricThe independent one-forms $d{x}^1,d{x}^2,d{x}^3,dt,du-2A_{i}d{x}^i-Udt$ exhibit three positive directions and a two-dimensional block with one positive and one negative direction. Thus the metric is nondegenerate, of signature $(4,1)$. Its coefficients do not depend on $u$, and $g_{uu}=0$, so $\xi ={\partial }_u$ is a null Killing vector field. Indeed $g_{Au}$ are constant, so ${\Gamma }^A{}_{Bu}=0$ for the Levi-Civita connection and $\xi$ is parallel.

The geodesic Lagrangian for an affine parameter $\lambda$ isThe cyclic coordinate $u$ gives the conserved quantity $p_u=\partial L/\partial u^{\prime }=t^{\prime }$. Work at a nonzero value of $p_u$ and rescale the affine parameter to set $t^{\prime }=1$. This is the essential step in the null Kaluza-Klein reduction of a stationary force: one fixes the momentum along the null isometry and projects its geodesics, rather than dividing by $g_{uu}$.

The spatial Euler-Lagrange equations, before setting $t^{\prime }=1$, areSince $t^{\prime \prime }=0$ and $F_{ij}{v}^j=-(\mathbf{B}\times \mathbf{v}{)}_i$, their reduction isEquivalently, after taking $t$ as the affine parameter, the term $\dot{u}$ is a total derivative and the reduced Lagrangian is $L_{\mathrm{red}}=\frac{1}{2}|\dot{\mathbf{x}}{|}^2-2\mathbf{A}\cdot \dot{\mathbf{x}}-U$. Its Euler-Lagrange equations give the same sign and factor two.

There is also an explicit converse using null geodesics. For any physical trajectory, setThis makes its five-dimensional tangent null. The conserved momentum of the cyclic coordinate $t$ becomesThe quantity in parentheses is conserved, because its derivative is $\dot{\mathbf{x}}\cdot (2\mathbf{B}\times \dot{\mathbf{x}})=0$. Hence the $t$ equation, as well as the spatial and $u$ Euler-Lagrange equations, is satisfied. Every trajectory admits a null geodesic lift with $p_u=1$, and every such lift projects to the required trajectory.

Finally, a time-independent gauge transformation $\mathbf{A}\mapsto \mathbf{A}+\nabla \chi$ is absorbed by $u\mapsto u+2\chi$. The one-form $du-2A_{i}d{x}^i-Udt$ and the five-dimensional metric are unchanged. These gauge transformations of an Eisenhart-Duval lift alter the reduced Lagrangian only by the total derivative $-2d\chi /dt$.

With the sign convention of the displayed operator, the Lax equation is $L_t=[L,A]$. Differentiating $L\phi =k^2\phi$ shows that ${\phi }_t+A\phi$ is another solution at the same spectral parameter. The normalization at $-\infty$ fixessince $A{e}^{-ikx}=4i{k}^3{e}^{-ikx}$ there. At $+\infty$, comparison of the two plane-wave coefficients givesThus the transmission denominator $a$ is time independent. This is the Time evolution of KdV scattering data for the stated normalization.

The logarithmic derivative $S={\phi }_x/\phi +ik$ obeys the Riccati equationIts formal large-$k$ expansion givesThere is an important qualification to the integral identity in the question. For real $k$ with nonzero reflection, $e^{ikx}\phi \sim a+b{e}^{2ikx}$ has no limit at $+\infty$, so the literal improper integral of the exact $S$ need not exist. Continue to $\mathrm{Im}k>0$, where the reflected exponential decays relative to $e^{-ikx}$, and take sufficiently large $k$ away from zeros and with a consistent logarithm. Then the normalization givesThe second equality is an asymptotic expansion, not an asserted convergent series. Time independence of $a$ fixes every coefficient, so each $\int S_{n}dx$ is a conserved quantity of the Korteweg-De Vries equation. The first two nontrivial examples are $-\int udx$ and $\int u^{2}dx$.

For the usual real-valued KdV potential and real formal $k$, write $S=X+iY$ with real coefficient functions. Its imaginary part of the Riccati equation is $Y^{\prime }-2kX+2XY=0$, henceIn the formal expansion, $X$ contains precisely the even-indexed $S_n$ and $Y$ the odd-indexed ones. Rapid decay of $u$ and its derivatives makes every coefficient of the logarithm vanish at both spatial ends. Therefore $\boxed{{\int }_{\mathbb{R}}{S}_{2j}dx=0(j\ge 1)}$. This is a coefficientwise formal identity; it does not claim that the exact real-axis reflected solution has a convergent integral of $X$. The analytically continued scattering identity and the formal expansion are the conventions needed to make the requested argument valid.

The Euler-Lagrange equations for the kinetic Lagrangian areExpanding and multiplying by the inverse metric gives the geodesic equationThe kinetic variational principle uses an affine parameter; the energy $g_{ab}{\dot{x}}^a{\dot{x}}^b/2$ is constant along its solutions. Constant curves are also geodesics.

For the Poincaré half-plane model, translation in $x$ gives the conserved quantity $p=\dot{x}/y^2$, and the energy is $E=({\dot{x}}^2+{\dot{y}}^2)/(2y^2)$. If $p=0$, the nonconstant geodesics are vertical lines, with $y=y_0{e}^{\pm \sqrt{2E}t}$. If $p\ne 0$,Writing $R^2=2E/p^2$ and integrating $dx/dy=\pm y/\sqrt{R^2-y^2}$ yields $(x-x_0{)}^2+y^2=R^2$. Thus nonconstant geodesic images are vertical lines or upper semicircles meeting the real axis orthogonally.

For the two specified points, the circle centre is $(6,0)$ and the radius is five:For example, an affine unit-speed parametrization is $x=6+5\tanh t$, $y=5\mathrm{sech}t$. The endpoints on the real axis are $(1,0)$ and $(11,0)$, outside the hyperbolic surface.

Let $f$ and $g$ be conserved quantities, allowing explicit time dependence. ThenUsing bilinearity of the Poisson bracket, commutation of partial differentiation with it, and the Jacobi identity for the Poisson bracket,Thus $\{f,g\}$ is also conserved.

If $g{f}^{\prime \prime }(a)>{\omega }^2$, defineThe linearized equation is $\ddot{ϵ}+{\nu }^2ϵ=0$, with bounded simple harmonic motionThus the strict stable equilibrium condition isThis is also a sufficient condition for local nonlinear Lyapunov stability. Indeed the equation for the bead on a uniformly rotating wire has the conserved quantityIts derivative is $\dot{r}$ times the governing equation moved to one side, so it vanishes. At $a$, the last two terms have zero first derivative and positive second derivative $g{f}^{\prime \prime }(a)-{\omega }^2$. Their strict local minimum traps sufficiently small perturbations. The motion is bounded rather than asymptotically returning to rest.

The Euler-Lagrange equations in cylindrical coordinates areThe last two follow because $\varphi$ and $z$ are cyclic coordinates. They give the conserved quantitiesSince the Lagrangian is autonomous, its energy function is conserved as well:The velocity-linear term cancels from this energy, but it still changes the equations of motion. A dimensionless logarithm means $\log (R/R_0)$ for some fixed reference length; replacing the printed $\log R$ by that expression changes the Lagrangian only by a total time derivative and shifts the numerical value of $k$, not the trajectory.

The coordinate $\theta$ is cyclic. Its conjugate momentum is therefore a second conserved quantity,It differs from mechanical angular momentum $m{r}^2\dot{\theta }$. The magnetic contribution is essential to the rotational Noether charge in this axisymmetric gauge.

The null space consists of concentration patterns on which transport produces no change. Here it is the familyEvery box has the same concentration. Ocean circulation may continue, but inflow and outflow of the chemical tracer balance in each box. With $\mathbf{b}=0$ these are exactly the steady states; physically admissible members have $\alpha \ge 0$. With nonzero $\mathbf{b}$, adding $\alpha 1$ to any particular steady state leaves its transport balance unchanged, but a null space vector alone need not balance the sources and sinks.

To identify the tracer inventory, apply conserved inventory from a left null vector. Directly summing each column givesConsequently, for the supplied ordinary differential equations,For $\mathbf{b}=0$, $M$ is a conserved quantity. The connected box model equalizes concentrations, approaching the unique member of the null space with the initial inventory:The arithmetic mean is appropriate here because the supplied matrix has zero column sums. In a general box model the conserved quantity is $w^T\mathbf{c}$, where $w$ is a positive left null space vector; for physical concentrations expressed per unit volume, these weights represent the box volumes up to a common factor. One must establish the weights from the equations, rather than infer equal volumes from a schematic drawing.

Nonnegative initial concentrations remain nonnegative under transport alone. Indeed $A+2I$ is a nonnegative matrix, sohas nonnegative entries for $t\ge 0$: it is a positive matrix semigroup. Its column sums are one by $1^{T}A=0$. To see why equalization occurs, set $d=\mathbf{c}-(M/4)1$ for $\mathbf{b}=0$; then $1^{T}d=0$ andThe right side vanishes only for constant $d$, which in the zero-sum vector subspace means $d=0$. Its strict negative bound on the unit sphere in that vector subspace gives decay to zero. Thus the matrix exponential redistributes inventory without creating negative concentrations; arbitrary prescribed sinks can still exhaust the tracer if they are allowed to act indefinitely.

Let a symmetric top have principal moments of inertia $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$. Apply a torque $({\tau }_1,{\tau }_2,0)$ constant in a principal body frame. Then the axial angular velocity $c={\omega }_3$ is a conserved quantity. Define $z={\omega }_1+i{\omega }_2$ and $\nu =(I_{\parallel }-I_{\perp })c/I_{\perp }$. The first two of Euler's equations combine into the equation in the title. This derivation uses constancy in the principal body frame, which generally does not mean constant torque in an inertial frame.

Multiplying by the integrating factor $e^{-i\nu t}$ proves, for $\nu \ne 0$,Equivalently $z$ rotates about the constant point $z_{\ast }=i({\tau }_1+i{\tau }_2)/(\nu I_{\perp })$ in the complex transverse plane. When $\nu =0$, the solution is instead $z(t)=z(0)+({\tau }_1+i{\tau }_2)t/I_{\perp }$. The angular velocity is $(\mathrm{Re}z,\mathrm{Im}z,c)$.

For a time-independent Hamiltonian $H$, the explicit time partial derivative of a smooth time-dependent conserved quantity is again a conserved quantity. Here conservation means the identity $D_{t}f=f_t+\{f,H\}=0$ on phase space, with the standard Poisson bracket. Differentiating this identity at fixed canonical coordinates givesBecause $H_t=0$, this is exactly $D_t(f_t)=0$, proving the assertion. Iteration gives the same result for all higher explicit time partial derivatives for which the differentiations are defined. By Hamiltonian energy balance, requiring $H$ itself to be a conserved quantity on all trajectories is equivalent to $H_t=0$.