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Let $M=M_1+M_2$ and $\alpha =M_2/M$, with $0<\alpha <1$. In the centre of mass frame the two primary distances are $r_1=\alpha R$ and $r_2=(1-\alpha )R$. Applying Newton's law of universal gravitation and circular centripetal acceleration to the first primary gives $M_1{\omega }^2\alpha R=G{M}_1{M}_2/R^2$, and henceThe same result follows from the second primary or from the relative two-body equation of motion.

A rotating-frame equilibrium has zero rotating-frame velocity, so its Coriolis force vanishes. Above the orbital plane both primaries have downward gravitational components; below it both have upward components. The centrifugal force has no $z$ component. More explicitly,which vanishes only at $z=0$ for finite nonsingular points. Thus all equilibria lie in the orbital plane.

The equation of motion in a rotating frame for the dwarf isOn the orbital plane the centrifugal contribution is $m{\omega }^2\mathbf{r}$. Thus the rotating-frame effective potential is $E=U-m{\omega }^2(x^2+y^2)/2$; equilibria are exactly its stationary points. The Coriolis force affects motion about those equilibria but not their positions. This is the circular restricted three-body problem, with the dwarf treated as a test mass.

The printed energy unit has the wrong dimensions. The correct total energy unit is $mGM/R$, or $GM/R$ for specific energy; $GM/R^2$ is an acceleration. Dividing $E$ by $mGM/R$ and measuring lengths in units of $R$ gives the dimensionless effective potentialThe primary coordinates are $(-\alpha ,0)$ and $(1-\alpha ,0)$. Restricting to $y=0$ gives $F(x)=\mathcal{E}(x,0)$ andThe absolute values are essential; dropping them would reverse the attraction on the left of a primary. On each of the intervals separated by the two primaries, $F^{\prime }$ decreases strictly from $+\infty$ to $-\infty$. There is exactly one Collinear Lagrange point on each interval, a maximum along the $x$ direction.

For the inner Lagrange point, write $x=1-\alpha -\delta$ with $\delta >0$. For the L2 Lagrange point, write $x=1-\alpha +\delta$. The respective equations areBalancing the leading terms gives $\delta =(\alpha /3{)}^{1/3}+O({\alpha }^{2/3})$. Therefore the small-mass-ratio collinear Lagrange points obeyThe omitted $-\alpha$ displacement of the secondary is smaller than the stated error. The leading distance from the secondary is the dimensionless Hill radius.

For the L3 Lagrange point put $x=-1+c\alpha +O({\alpha }^2)$. It lies to the left of both primaries, soExpanding the two positive inverse-square factors gives $(1-\alpha )/(x+\alpha {)}^2=1+(2c+1)\alpha +O({\alpha }^2)$ and $\alpha /(x-1+\alpha {)}^2=\alpha /4+O({\alpha }^2)$. Thus $F^{\prime }=-(3c+5/4)\alpha +O({\alpha }^2)$, soAll three have $y=z=0$. The small-$\alpha$ equalities in the paper are asymptotic expressions, not exact coordinates for finite mass ratio.

For the remaining Lagrange points, introduce plane polar coordinates about the first primary: $x=-\alpha +s\cos \theta$, $y=s\sin \theta$. Its distance to the second primary is $d=(s^2+1-2s\cos \theta {)}^{1/2}$, andThe angular stationary condition is ${\partial }_{\theta }\mathcal{E}=\alpha s\sin \theta (d^{-3}-1)=0$. For an off-axis point $s>0$ and $\sin \theta \ne 0$, so $d=1$. The radial stationary condition becomesHence $s=1$, and $d=1$ implies $\cos \theta =1/2$. The Triangular Lagrange points are exactlyBoth triangles are equilateral. The collinear and off-axis calculations together exhaust the stationary points.

For the requested sketch, $\mathcal{E}$ tends to $-\infty$ at either primary and at large in-plane radius. Each Collinear Lagrange point is a saddle of the planar effective potential: its $x$ curvature is negative and its $y$ curvature is positive. To see the latter, let $d_1=|x+\alpha |$, $d_2=|x-1+\alpha |$ and $D=(1-\alpha )/d_1^3+\alpha /d_2^3$. At an inner point both distances are below one, so $D>1$. At an exterior stationary point, use barycentric positions $a_1=-\alpha$, $a_2=1-\alpha$ and weights $1-\alpha ,\alpha$: the equation implies $x(D-1)=\sum w_i{a}_i/|x-a_i{|}^3$. To the right, the weighted sum is positive because the positive primary is closer; to the left it is negative because the negative primary is closer. Thus $D>1$ there as well, and ${\mathcal{E}}_{yy}=D-1>0$.

At either Triangular Lagrange point, the planar Hessian isIts trace is $-3$ and determinant $27\alpha (1-\alpha )/4>0$, so these are local maxima. Their height is $\mathcal{E}=-3/2+\alpha (1-\alpha )/2$. The sketch below uses a finite mass ratio to separate the five points clearly; its singular wells are clipped only for display. These curvature labels concern the effective potential, not a stability test that ignores the Coriolis force.

The displayed estimate has two independent defects. Even for $a(v)=v$, its right side cannot use $f(t)$. At $t=1$, choose nonnegative smooth functions $\psi ,\chi$ with compact support and prescribe $f(1,x,v)=\psi (x)\chi (v)$. This is a legitimate transported solution with $f_{\mathrm{in}}(x,v)=\psi (x+v)\chi (v)$. The proposed inequality would requireVelocity dilation makes $\Vert \chi {\Vert }_1/\Vert \chi {\Vert }_{\infty }$ arbitrarily large, while the other factors stay fixed. Thus no universal $C_a$ works with the same-time mixed Lebesgue norm.

There is a second issue after replacing $f(t)$ by $f_{\mathrm{in}}$: the Jacobian determinant bounds give a local diffeomorphism, not necessarily a one-to-one map. Suppose additionally that $a$ is injective. With $y=x-ta(v)$, the change of variables formula and $|\det Da|\ge {\alpha }_1$ giveSurjectivity is not needed because the domain of integration can be enlarged. With injectivity, the corrected estimate has $C_a=1/{\alpha }_1$. The upper bound ${\alpha }_2$ is unnecessary.

More generally, dispersion with a nonlinear velocity map uses the area formula to sum over inverse branches. Define the weighted inverse multiplicityIf ${\mathrm{esssup}}_w{m}_a(w)<\infty$, the same argument gives the corrected estimate with $C_a=\Vert m_a{\Vert }_{\infty }$. At most $N$ inverse branches give $C_a\le N/{\alpha }_1$. Thus the missing global assumption concerns multiplicity, not merely local volume distortion.

For a noninjective map with constant Jacobian determinant giving a counterexample in two dimensions, write $v=(s,r)$ and setThe polar coordinates calculation gives $\det Da=e^{2s}{e}^{-2s}=1$, yet $a(0,2\pi k)=(1,0)$ for every integer $k$. To turn this into a failure of the estimate, take a small open disk $W$ about $(1,0)$ that avoids the origin. For each of $N$ inverse branches over $W$, choose a smooth cutoff ${\chi }_k(v)=\chi (a(v))$ supported on that branch, where $\chi$ has support strictly inside $W$ and is extended by zero. These velocity supports are disjoint. SetThen $\Vert f_{\mathrm{in}}{\Vert }_{L_x^1{L}_v^{\infty }}=\Vert \psi {\Vert }_1\Vert \chi {\Vert }_{\infty }$, independent of $N$, whereas at a fixed $t>0$ and suitable $x$,The last integral is positive and independent of $N$. No finite $C_a$ exists for this fixed $a$, although ${\alpha }_1={\alpha }_2=1$. In one dimension, by contrast, a nonvanishing derivative has a constant sign, so the lower derivative bound makes $a$ a global diffeomorphism.

Use the Fourier transform convention $\hat{\nu }(\xi )=\int e^{-ix\cdot \xi }d\nu (x)$ throughout this question. The decay assumption implies square integrability, since polar coordinates giveNear zero the integrand is bounded by $r$, and at infinity it is bounded by $r^{-1-2ϵ}$. The positive $ϵ$ is what makes the latter integrable.

By the Plancherel theorem, there is a function $g\in L^2({\mathbb{R}}^2)$ whose Fourier transform is $\hat{\mu }$. It is the density of $\mu$ with respect to Lebesgue measure. To justify this step rather than assume a density, for every Schwartz function $\phi$, Fourier inversion givesThe finite measure and the locally integrable function thus define the same tempered distribution, so they agree as measures: $d\mu =gdx$. In particular $g\ge 0$ almost everywhere and $\int g=\mu ({\mathbb{R}}^2)<\infty$. This is the L2 density from a square-integrable Fourier transform principle.

For $f\in L^{\infty }(\mu )$, the density of $fd\mu$ is $fg$. The inequality $|f|\le \Vert f{\Vert }_{L^{\infty }(\mu )}$ holds wherever $g>0$, apart from a Lebesgue measure zero set, soA second application of the Plancherel theorem yields the explicit boundThe implicit constant in the requested estimate may depend on the measure and its Fourier-decay bound, but is independent of $f$.

An embedded surface parametrization is a smooth map $F:U\subset {\mathbb{R}}^2\to {\mathbb{R}}^3$, with open $U$, injective differential of rank two at each point, and $F$ a homeomorphism onto its image with the subspace topology. Its induced metric isIn coordinates, an arc-length-parametrized geodesic satisfieswhere the Christoffel symbols are ${\Gamma }_{ij}^k=\frac{1}{2}{g}^{k\ell }({\partial }_i{g}_{j\ell }+{\partial }_j{g}_{i\ell }-{\partial }_{\ell }{g}_{ij})$. The normalization fixes unit speed; it is compatible with the differential equations because geodesic speed is constant.

For the cone use $F(r,\theta )=(r\cos \theta ,r\sin \theta ,\sqrt{3}r)$, $r>0$, restricting $\theta$ to a suitable open interval for a chart. Direct differentiation gives its first fundamental formWith $\rho =2r$ and $\varphi =\theta /2$, this becomes $d{\rho }^2+{\rho }^{2}d{\varphi }^2$, the Euclidean metric in polar coordinates. The map $(r,\theta )\mapsto (2r\cos (\theta /2),2r\sin (\theta /2))$ is therefore a local isometry. It is local rather than global because increasing $\theta$ by $2\pi$ increases $\varphi$ by $\pi$.

Given two points, choose their angle representatives so that $|{\theta }_2-{\theta }_1|\le \pi$. Their developed plane angles then differ by at most $\pi /2$. The straight segment between their developed positions stays away from the origin: it lies in an angular sector of width at most $\pi /2$ containing both positive-radius endpoints. Develop that segment back onto the cone. Since local isometries preserve geodesics, it gives a geodesic joining the two points; it can be parametrized by arc length. This constructs geodesics on a punctured circular cone. For a repeated point the constant curve is the degenerate geodesic. In particular, this argument does not appeal to completeness of the punctured cone.

The joining geodesic need not be unique. For endpoints with equal positive radius and opposite azimuth, one may use the angular changes $+\pi$ and $-\pi$. In the plane these give two straight segments from $(\rho ,0)$ to $(0,\rho )$ and $(0,-\rho )$, respectively. Both avoid the origin. Their lifts pass on opposite sides of the cone and have distinct images, so a reparametrization cannot identify them.

A bounded-domain hypothesis is needed for the requested global construction and uniqueness. The printed question does not include it. For example, the upper half-plane satisfies the stated exterior cone condition, but both $u=0$ and $u(x_1,x_2)=x_2$ are harmonic functions with zero Dirichlet boundary condition. Thus part (iv), in its printed unrestricted solution class, is false. In this question's solutions we add that $\Omega$ is bounded; the exterior cone condition and all other data remain as printed.

The half-plane also rules out the printed global separated barrier. In its angular interval $(0,\pi )$, varying $r$ on a fixed ray forces $\nu =2$ if $r^{\nu -2}(h^{\prime \prime }+{\nu }^{2}h)\ge \delta >0$ for every $r>0$. We would then have $h^{\prime \prime }+4h\ge \delta$ and $h<0$. Put $a=\pi /4$ and $b=3\pi /4$. Twice integration by parts givescontradicting the positive integrand. Thus a source qualification is necessary for part (i) as well as part (iv).

Fix $x_0$ and its exterior cone of half-angle $\alpha$. Choose $0<{\alpha }_0<\min (\alpha ,\pi /2)$ and setOn the complement of the cone, unwrap the angle in polar coordinates as $\vartheta \in (\alpha ,2\pi -\alpha )$, measured from the cone axis. Then $\psi =\vartheta -\pi$ is the angle from the opposite axis and $|\psi |\le \pi -\alpha$ on $\overline{\Omega }\setminus \{x_0\}$. Define the power barrier for an exterior cone byThis has the requested separated form $r^{\nu }h(\vartheta )$. In the printed signed-angle convention, the same function is $-r^{\nu }\cos (\beta (\pi -|\theta |))$ outside the cone. It is smooth across the negative axis: near that axis the angular expression is the even function $\cos (\beta \psi )$.

Since $\beta (\pi -\alpha )<\pi /2$, put $m=\cos (\beta (\pi -\alpha ))>0$. The cosine is at least $m$, so $g\le -m{r}^{\nu }<0$ away from $x_0$, and $g$ is continuous at $x_0$. The Laplacian in polar coordinates givesLet $D={\sup }_{x\in \Omega }|x-x_0|<\infty$. Since $\nu <2$, the barrier for the Dirichlet problem satisfiesBoth the uniform lower bound and the later global comparison use boundedness. In particular, $-g$ is positive at every other boundary point, and is bounded away from zero on boundary sets staying a positive distance from $x_0$.

For $r>0$, differentiate $r^2=x^2+y^2$ and use the angular derivative $(x\dot{y}-y\dot{x})/r^2$. The polar coordinates satisfyHence the rotation is clockwise. The origin is an unstable equilibrium, while $r=1$ is a limit cycle of period $2\pi$. Writing $u=r^2$ gives the logistic differential equation $\dot{u}=2u(1-u)$. For $r(t_0)=r_0>0$,All nonzero solutions tend to the unit-circle attracting limit cycle as $t\to \infty$, spiralling outward for $r_0<1$ and inward for $r_0>1$. For $0<r_0<1$ the solution exists for every real time and approaches the origin as $t\to -\infty$; for $r_0=1$ it remains periodic in both time directions. For $r_0>1$, the denominator first vanishes atThis finite-time blow-up of an ordinary differential equation occurs backward in time. Such exterior solutions have no $t\to -\infty$ behavior: their maximal time interval is $(t_{\ast },\infty )$. The zero solution exists for all time and has no defined polar angle.

A central force has no component in the angular direction. The angular component of the acceleration in polar coordinates therefore gives $2\dot{r}\dot{\theta }+r\ddot{\theta }=0$, orFor unit mass, $h=r^2\dot{\theta }$ is the signed angular momentum perpendicular to the orbital plane, so its magnitude $|h|$ is conserved. Equivalently, the torque $\mathbf{r}\times \mathbf{F}$ vanishes, which also fixes that plane.

The given potential energy is $k/r$, whose radial force is $k/r^2$: positive $k$ is repulsive and negative $k$ attractive. The kinetic energy in polar coordinates is $({\dot{r}}^2+r^2{\dot{\theta }}^2)/2$, so conservation of energy givesFor $h\ne 0$ and $k>0$, the effective potential is positive, strictly decreasing from $+\infty$ to $0$, and has no stationary point. For $k=-\kappa <0$, it still tends to $+\infty$ at zero, crosses zero at $r=h^2/(2\kappa )$, and has its unique minimum atIt then approaches zero from below. The minimum corresponds to a stable circular orbit. If $h=0$, the centrifugal barrier disappears; the attractive effective potential is simply $-\kappa /r$ and has no minimum.

Interpret projection from very far away as the asymptotic incoming state at $r\to \infty$. For $v>0$ and impact parameter $b>0$, conservation of energy and conservation of angular momentum give $E=v^2/2$ and $|h|=bv$. At the closest approach the radial velocity vanishes. With $k=-\kappa <0$, the central-force radial turning point equation isOnly the positive root is physical. Put $q=\kappa /v^2$. The pericentre distance and the purely tangential speed there areThus $p<b$ and $u>v$: attraction bends the path inward and increases the speed. With positive energy, the Kepler orbit is a hyperbola. Measured from the pericentre direction, its polar coordinates satisfy $r=h^2/[\kappa (1+e\cos \varphi )]$ with $e=\sqrt{1+b^2{v}^4/{\kappa }^2}>1$.

The hyperbola shown has a nonzero impact parameter. For $b=0$, the attractive radial orbit instead reaches the singular origin; there is no regular turning point with a finite closest-approach speed.

Put $\mathbf{t}=\mathbf{\theta }$. The product rule applied to $Q_{ij}=\lambda (t_i{t}_j-{\delta }_{ij}/2)$ givesSince $\mathrm{Tr}{Q}^2={\lambda }^2/2$, substituting this vector into the divergence elastic term proves the requested form of the local density. For the radial amplitude and tangential director, the polar coordinates identities areHence the tangential integer nematic defect core hasThe $1/r$ director-curvature term survives when the amplitude is constant. Replacing the divergence norm by a full tensor-gradient norm would not give this expression; it is the PDF's divergence convention that is being evaluated.

The equilibrium equations are $x(-1+x^2+\beta y^2)=y(-1+\beta x^2+y^2)=0$. For $\beta \ne 1$ the fixed points are the origin, four axis points $(\pm 1,0),(0,\pm 1)$, and four diagonal points $(\pm s,\pm s)$ with independent signs and $s=(1+\beta {)}^{-1/2}$. The hypothesis $\beta >-1$ ensures that $s$ is real. The Jacobian matrix has diagonal entries $-1+3x^2+\beta y^2$, $-1+\beta x^2+3y^2$ and off-diagonal entries $2\beta xy$.

At the origin both eigenvalues are $-1$, so it is a stable node. At each axis point the eigenvalues are $2,\beta -1$. At each diagonal point they are $2,2(1-\beta )/(1+\beta )$, along the radial and transverse diagonal directions respectively. ThusThe axes and diagonals are invariant lines. For $\beta =1/2$ the axis saddle points have stable directions transverse to their axes and unstable directions along them; the diagonal points are unstable nodes. For $\beta =2$ these roles switch: the diagonal saddle points have stable transverse directions and unstable radial directions, and the axis points are unstable nodes. Their stable manifolds separate trajectories attracted to the origin from outward-escaping trajectories. Symmetry gives the other quadrants.

For $\beta =1$, the entire unit circle consists of equilibria. In polar coordinates $\dot{r}=r(r^2-1)$ and $\dot{\theta }=0$: every ray inside the circle moves toward the origin, and every ray outside moves outward. The circle has neutral tangential and unstable radial directions; its points are not isolated saddle points or unstable nodes.

All three phase portraits are shown below. The potential $\Phi =-(x^2+y^2)/2+(x^4+y^4)/4+\beta x^2{y}^2/2$ satisfies $(\dot{x},\dot{y})=\nabla \Phi$ and $\dot{\Phi }=|\nabla \Phi {|}^2$, ruling out nonconstant periodic orbits. Blue curves show the saddle stable separatrices, integrated backwards from their local stable directions. The flow arrows and invariant lines, together with the equilibrium classification, specify the topology of the sketches.

For a one-to-one differentiable change of variables $(x,y)\mapsto (u,v)$, the change of variables formula iswhere

Here $u=x+y$, $v=x^2+y^2$, andAt fixed $v$, write $x=\sqrt{v}\cos \theta$, $y=\sqrt{v}\sin \theta$ with $0\le \theta \le \pi$. Then $u=\sqrt{v}(\cos \theta +\sin \theta )$, soThe constant-$v$ curves are semicircles and the constant-$u$ curves are lines $x+y=u$.

The map is two-to-one where $\sqrt{v}<u<\sqrt{2v}$, because both $(x,y)$ and $(y,x)$ lie in the upper half-disc, and one-to-one elsewhere. Since $|y-x|=\sqrt{2v-u^2}$, accounting for this multiplicity givesDirectly, symmetry makes ${\iint }_{D}xdA=0$, while polar coordinates give ${\iint }_{D}ydA={\int }_0^1{r}^{2}dr{\int }_0^{\pi }\sin \theta d\theta =2/3$.

Define $T^{†}(0)=0$; the printed quotient is undefined at the origin, but this continuous extension changes no transport cost because $\mu (\{0\})=0$. In polar coordinates, the two probability density functions giveBoth angular distributions are uniform, with independence of angle and radius. The proposed transport map preserves the angle and sends $r$ to $s=\sqrt{r}$. Thereforeand preservation of the angle proves $(T^{†}{)}_{\#}\mu =\nu$. As a local check using the Jacobian determinant, its radial and tangential derivatives for $r>0$ are $1/(2\sqrt{r})$ and $1/\sqrt{r}$, respectively, so $\det D{T}^{†}=1/(2r)$ and $g(T^{†}(x))\det D{T}^{†}(x)=(2r/\pi )/(2r)=f(x)$.

Now use the convex functionIt is convex because the Euclidean norm is convex and $s\mapsto (2/3)s^{3/2}$ is increasing and convex on $[0,\infty )$. It is differentiable, including at zero, andIts graph transport plan lies in the graph of $\partial \Phi$, so the Knott–Smith optimality criterion proves quadratic optimality. Since that plan is induced by a map, part 1(b) proves optimality for the Monge optimal transport problem as well. The minimum provides a useful independent check: