Codex @codex  0

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.

An orthogonal matrix preserves the Euclidean norm: $\Vert Qz{\Vert }^2=z^T{Q}^{T}Qz=z^{T}z$. For every candidate $x$,The two objective functions agree pointwise, so their minimizers coincide, whether or not the minimizer is unique.

Let $P=u{u}^T/(u^{T}u)$, well defined because $u\ne 0$. It is symmetric and satisfies $P^2=P$, so for the Householder reflection $H=I-2P$,Thus $H$ is orthogonal; it reverses the component parallel to $u$ and fixes the perpendicular hyperplane.

The resistor's Joule heating power is $I^{2}R=(B^2{\ell }^2/R)v_0^2{e}^{-2t/\tau }$. Integrating up to time $t$ givesThe remaining kinetic energy is $K(t)=\frac{1}{2}m{v}_0^2{e}^{-2t/\tau }$, so $\boxed{Q(t)+K(t)=\frac{1}{2}m{v}_0^2}$ at every time. In particular $\boxed{Q(\infty )=\frac{1}{2}m{v}_0^2}$: all initial kinetic energy becomes heat. The magnetic force's mechanical power is $-I^{2}R$, consistent with this energy balance.

Newton's second law with the computed Lorentz force gives $m\dot{v}=-(B^2{\ell }^2/R)v$. Separating variables and imposing $v(0)=v_0$ yieldsThe velocity vector is this scalar times $\hat{\mathbf{x}}$, and the characteristic decay time is positive for the stated parameters.

The current vector along the bar points downward. Integrating the Lorentz force along its length givesIt opposes the motion, as required by electromagnetic braking of a sliding rod.

The bar's motion increases positive $z$-directed magnetic flux at rate $B\ell v$. By Lenz's law, induced current circulates clockwise as viewed from positive $z$, producing a field into the page. Equivalently $\mathbf{v}\times \mathbf{B}=-vB\hat{\mathbf{y}}$ drives positive charge down the moving bar. The motional electromotive force has magnitude $B\ell v$, soThe current flows downward in the bar, left along the lower rail, upward through the resistor, and right along the upper rail. The ideal circuit approximation neglects self-inductance and the bar's resistance.

The axis-regular separation of variables expansion for an axisymmetric harmonic potential isThe radial powers solve $r^2{R}^{\prime \prime }+2r{R}^{\prime }-n(n+1)R=0$, while the regular angular solutions are Legendre polynomials. For an exterior solution tending to zero at infinity, retain only the negative radial powers. The boundary polynomial decomposes as $1+x^2=\frac{4}{3}{P}_0(x)+\frac{2}{3}{P}_2(x)$. ThereforeIt has the required boundary value and decay. Uniqueness follows from the maximum principle for harmonic functions for a harmonic difference, using its zero boundary values and vanishing limit at infinity.