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The coordinate curves are assumed to be geodesics as geometric curves; use their arc-length parametrization in the geodesic equations. On $u=c$, choose increasing arc length $s$, so $\dot{u}=0$ and $\dot{v}=1/\sqrt{G}$. The transverse equation becomesThus $(F/\sqrt{G}{)}_v=G_u/(2\sqrt{G})=(\sqrt{G}{)}_u$. On $v=d$, similarly $\dot{v}=0$, $\dot{u}=1/\sqrt{E}$, and the transverse equation gives $(F/\sqrt{E}{)}_u=E_v/(2\sqrt{E})=(\sqrt{E}{)}_v$. Hence the geodesic coordinate grid identities are

For $t>0$, use the Bromwich inversion formula with a vertical contour $\mathrm{Re}s=c>0$:Close the contour to the left, where the exponential decays. The large-arc contribution tends to zero by the usual exponential arc estimate, and the two simple poles are at $s=ia$ and $s=-ia$. The residues of the complete integrand are $e^{iat}/2$ and $e^{-iat}/2$. The residue theorem therefore givesThe right limit at zero is one. The sign of nonzero real $a$ has no effect on the answer.

The set is a rational comb with a connecting spine: it contains the whole vertical segment $x=0$, and every horizontal segment at rational height. Any point with $x>0$ has rational height and can move horizontally to the vertical segment without leaving the set. It can then move vertically to $(0,0)$. Points already on the vertical segment use only the second step. Concatenating these polygonal paths connects any two points. ConsequentlyNo claim about a horizontal segment at irrational height is needed: at such a height, the only point in the set lies on the connecting spine.

Use the continuous slope map $(x,y)\mapsto x/y$ on the given set, where $y>0$. The attained ratios are positive rational numbers. Choosing the irrational cut $\sqrt{2}$, the two subsetsare disjoint, relatively open, and cover the entire set. Both are nonempty: ratios one and two are attained inside the square. This irrational-cut separation of rational slopes is a separation, soThe individual rational-slope rays are connected, but their common origin is excluded; their separate connectivity does not imply connectivity of their union.

Substitute the rotating velocity field into the Euler equations for an inviscid fluid. Its acceleration is $(-\dot{\Omega }y-{\Omega }^{2}x,\dot{\Omega }x-{\Omega }^{2}y,0)$, soThe preceding angular acceleration makes both mixed coefficients $(\beta +\gamma )/2$, ensuring integrability. ThusPressure is determined up to the spatially uniform function $p_0(t)$. The specified force has no vertical component; if gravity were additionally included in the body force, the usual hydrostatic term $-\rho gz$ would also appear. It is not implied by the displayed force here.

The Euler equations for an inviscid fluid are ${\partial }_t\mathbf{v}+(\mathbf{v}\cdot \nabla )\mathbf{v}=-\nabla p/\rho +\mathbf{f}$. Here the vorticity is $2\Omega {\mathbf{e}}_z$, and direct differentiation gives $(\mathbf{v}\cdot \nabla )\mathbf{v}=-{\Omega }^2(x,y,0)$, whose curl vanishes. Taking the curl therefore givessoFor constant force coefficients and the initial rest condition, $\Omega (t)=(\gamma -\beta )t/2$. Only the antisymmetric part of the horizontal force gradient generates rotation; its symmetric part is balanced by pressure.

At height $z$, $\mathbf{r}\times \mathbf{v}=(-\Omega zx,-\Omega zy,\Omega (x^2+y^2))$. Integrating over a circular cross-section cancels the horizontal components by symmetry. Thus the angular momentum per unit length isThe coefficient is the polar moment of inertia per unit length of the liquid column.

The rigid-rotation velocity field is $\mathbf{v}=(-\Omega y,\Omega x,0)$, with spatially uniform $\Omega (t)$. Its divergence is zero, since ${\partial }_x(-\Omega y)={\partial }_y(\Omega x)=0$. It therefore satisfies the incompressible continuity equation. On the cylindrical wall, $\mathbf{v}\cdot (x,y,0)=0$, so it has zero normal velocity; on either end cap its vertical component is zero. These are the impermeability boundary conditions for an inviscid flow. Tangential slip is permitted; a no-slip condition is not part of this inviscid model.

Use real physical fields in the quadratic energy expressions. The vector identity $\nabla \cdot (\mathbf{E}\times \mathbf{B})=\mathbf{B}\cdot (\nabla \times \mathbf{E})-\mathbf{E}\cdot (\nabla \times \mathbf{B})$ and vacuum Maxwell's equations giveIntegrating over a fixed volume and applying the divergence theorem proves Poynting's theorem:The left side is outward energy flux through the closed surface, not merely local intensity.

For a linearly polarized wave with phase $\chi =\mathbf{k}\cdot \mathbf{r}-\omega t$, the real fields give Poynting vector $\mathbf{S}={\epsilon }_{0}c{E}_0^2{\cos }^2\chi \hat{\mathbf{k}}$. Local intensity oscillates at $2\omega$. In the closed-surface integral, the time-independent part cancels because ${\oint }_A\hat{\mathbf{k}}\cdot d\mathbf{A}=0$. The remaining flux generally oscillates at $2\omega$, has zero time average, and may vanish identically for special volume geometries.

For circular polarization, take $\mathbf{E}=E_0({\mathbf{e}}_1\cos \chi +{\mathbf{e}}_2\sin \chi )$ with orthonormal transverse vectors. Then $E^2=E_0^2$ and $\mathbf{B}=\hat{\mathbf{k}}\times \mathbf{E}/c$, so $\mathbf{S}={\epsilon }_{0}c{E}_0^2\hat{\mathbf{k}}$ is spatially and temporally constant. Therefore the closed-surface Poynting flux of a plane wave isThere is still a nonzero steady local energy flow: equal energy enters and leaves every fixed closed volume. Calling the net closed-surface flux a nonzero constant would confuse these two quantities.

Vacuum Maxwell's equations are invariant under spatial rotations. Rotate the preceding coordinate system so that ${\mathbf{e}}_z$ becomes the propagation direction $\hat{\mathbf{k}}$, and rotate the transverse electric direction to a unit polarization vector $\hat{\mathbf{n}}$. Dot and cross products transform covariantly under this rotation, so transversality and the field relation are unchanged. With $\omega =c|\mathbf{k}|$,These complex fields encode real fields through their real parts. The displayed form describes linear electromagnetic polarization; a general transverse complex amplitude can describe elliptical or circular electromagnetic polarization as well.

For these plane waves, the vacuum Faraday law readsCancel the nonzero factor $i\omega$ to obtainFor positive-frequency propagation in the positive $z$ direction, $k/\omega =1/c$.

Substitute the monochromatic plane waves into the two divergence equations of Maxwell's equations. They become $ik{E}_{0z}=0$ and $ik{B}_{0z}=0$. For nonzero wavenumber,Thus both fields are transverse to the direction of propagation. The wave equation also gives the vacuum dispersion relation ${\omega }^2=c^2{k}^2$.

Take the curl of the vacuum Faraday law and use the vacuum Ampere-Maxwell law:Since the electric divergence vanishes, the double-curl identity makes the left side $-\Delta \mathbf{E}$. Therefore the electromagnetic wave equation isThe same calculation interchanging the fields gives the wave equation for $\mathbf{B}$.

Suppose an open hyperbolic disc centred at $X$ is contained in a hyperbolic triangle. Move $X$ to zero by an isometry. Extend the radial geodesics from zero through the three vertices to ideal points $P_1,P_2,P_3$. Because zero lies inside the original triangle, the three vertex directions are not contained in a semicircle. The ideal hyperbolic triangle with those endpoints therefore contains zero: otherwise a separating geodesic through zero would put all three endpoints in one semicircle. It also contains each finite vertex, which lies on the segment from zero to the corresponding ideal endpoint. Geodesic convexity then makes it contain the original triangle.

The three angular gaps between successive ideal endpoints sum to $2\pi$, and each is at most $\pi$. One has size at least $2\pi /3$, so its half-angle $\alpha$ is at least $\pi /3$. The distance in part (iii) decreases as $\alpha$ increases. Thus the corresponding side of the ideal triangle is at distance at mostfrom zero. If the disc radius exceeds this distance, it contains a point on that side and points beyond it, and cannot be contained in the ideal triangle, hence cannot be contained in the original one. This proves the hyperbolic triangle inradius bound:The symmetric ideal triangle has equality, explaining the constant. Finite triangles have strictly smaller inradius.

Rotate the disc so that $P=e^{i\alpha }$ and $Q=e^{-i\alpha }$, with $0<\alpha \le \pi /2$ representing half the smaller angle. For $\alpha <\pi /2$, the hyperbolic line joining them is the arc of a circle orthogonal to the unit circle. Its Euclidean centre is real, say $c$, and orthogonality gives squared radius $c^2-1$. Substituting $P$ into $|P-c{|}^2=c^2-1$ gives $c=\sec \alpha$, with radius $\tan \alpha$.

The arc's point nearest zero is on the positive real axis, atDistance from zero in the Poincare disc model increases with Euclidean modulus, by the radial-length calculation. ThereforeFor $\alpha =\pi /2$, the hyperbolic line is a diameter through zero and its distance is zero, agreeing with the continuous limit of the formula. Choosing the smaller endpoint angle avoids a spurious negative value caused by an inappropriate oriented-angle convention.

For the Poincare disc model, radial distance from zero to Euclidean radius $r$ isIt is minimal: any curve has length at least the integral of $2|dr|/(1-r^2)$, which in turn is at least the net radial integral. Consequently a hyperbolic circle of radius $a$ centred at zero is $|\zeta |=\tanh (a/2)$, a Euclidean circle.

Given a centre $p$, use a Poincare disc automorphism carrying $p$ to zero. Since it is an isometry, it takes the given hyperbolic circle to the preceding Euclidean circle. A Möbius transformation takes generalized circles to generalized circles. The preimage here is compact and contained in the disc, so it cannot be a straight line. ThusThis argument does not say that their Euclidean and hyperbolic centres coincide.

For $|\lambda |<1$, let $F_{\lambda }(\zeta )=(\zeta -\lambda )/(1-\overline{\lambda }\zeta )$. Direct calculation givesIts denominator does not vanish in the disc, and this identity shows that it maps the disc into itself. Its inverse $(w+\lambda )/(1+\overline{\lambda }w)$ has the same property, so $F_{\lambda }$ is a Poincare disc automorphism. Multiplication by a complex number of modulus one is another such automorphism.

Conversely let $T$ be a Möbius transformation mapping the disc onto itself, and let $\lambda =T^{-1}(0)$. Then $T\circ F_{\lambda }^{-1}$ is a disc automorphism fixing zero, say $S(\zeta )=a\zeta /(c\zeta +d)$ with $d\ne 0$. Since it maps the boundary circle to itself, $|a{|}^2=|c\zeta +d{|}^2=|c{|}^2+|d{|}^2+2\mathrm{Re}(c\overline{d}\zeta )$ for every $|\zeta |=1$. This forces $c\overline{d}=0$, hence $c=0$ and $|a/d|=1$. Thus $S$ is a rotation $\eta \zeta$, and $T=\eta F_{\lambda }$. Absorb the denominator's minus sign into $\omega =-\eta$ to obtain precisely the paper's convention:Here mapping the disc to itself means bijectively onto itself; there are nonautomorphic Möbius transformations whose image is a proper subdisc.

Because the velocity is independent of $z$, the vortex-stretching term is $(\mathbf{\omega }\cdot \nabla )\mathbf{u}={\omega }_z{\partial }_z\mathbf{u}=0$. Substituting ${\omega }_z=-{\nabla }^2\psi$, $u={\psi }_y$ and $v=-{\psi }_x$ into the vorticity equation and cancelling a minus sign givesThe planar vorticity is transported by the flow without vortex stretching under these inviscid assumptions.

For the stated incompressible planar flow, use the stream function convention $u={\psi }_y$, $v=-{\psi }_x$. It gives $u_x+v_y=0$ and the vorticity isThus $\boxed{\mathbf{\omega }=-{\nabla }^2\psi \hat{\mathbf{z}}}$, where the Laplacian is the two-dimensional one.

The first column already has zeros below the top entry. In the last three rows of the second column, reflect $w=(4,2,4{)}^T$ to $(-6,0,0{)}^T$ using $u=w+6e_1=(10,2,4{)}^T$. Since $u^{T}u=120$, the Householder reflection isWith $Q=\mathrm{diag}(1,H)$, direct multiplication givesThe transformed least-squares objective is $(x_1+2x_2-1{)}^2+(-6x_2+2/3{)}^2+50/9$. Its first two squares can both vanish, so the unique answer isUniqueness follows because the triangular leading block is invertible; this calculation implements Householder QR decomposition without squaring the condition number through normal equations.