Let $e_{ij}=({\partial }_i{u}_j+{\partial }_j{u}_i)/2$ be the rate-of-strain tensor. A Newtonian fluid has a linear isotropic constitutive law for its viscous stress tensor. To determine its form, diagonalize the symmetric tensor $e$. Rotations by $\pi$ about its principal axes leave $e$ fixed, forcing its stress response to be diagonal as well. Linearity and symmetry under permutations of the axes force each diagonal entry to be $\lambda \mathrm{tr}e+2\mu e_{ii}$, with the same constants for each axis. Rotating back gives ${\tau }_{ij}=\lambda e_{kk}{\delta }_{ij}+2\mu e_{ij}$. Since incompressibility means $e_{kk}=0$, adding the arbitrary isotropic pressure givesAt a material interface between ordinary viscous fluids without interfacial slip, velocity is continuous, and its normal component equals the interface's normal speed. Tangential traction is continuous when surface tension is constant. Normal traction has the curvature-dependent surface tension jump; in a convention defining signed curvature $\kappa$ by the balance, $({\sigma }_{+}-{\sigma }_{-})n=\gamma \kappa n$. With no surface tension, the full traction is continuous. At rigid stationary walls the no-slip boundary condition imposes zero velocity.

Put $G=-dp/dx>0$ and seek $u=u_{\pm }(y)e_x$. The convective acceleration vanishes, and the Navier-Stokes equations reduce to ${\mu }_{\pm }{u}_{\pm }^{\prime \prime }=-G$, with $p=p_0-Gx$. continuity of velocity and tangential traction at $y=0$ gives a common interface velocity $U_0$ and a common shear integration constant $C$:Imposing $u_{-}(-a)=u_{+}(a)=0$ givesThe velocity has no normal component, so the flat interface remains stationary, and its normal stresses match because pressure is common. These expressions give the two-layer plane Poiseuille flow with unequal viscosities.

Euler characteristic for a finite connected plane graph is $v-e+f=2$. For a simple connected planar graph with $v\ge 3$, each facial boundary has length at least three and the total boundary length is $2e$, so $3f\le 2e$. Combining gives $e\le 3v-6$, and therefore average degree of a graph $2e/v<6$. Some vertex has degree at most five. Components of size one or two have this property directly, so every nonempty finite simple planar graph has $\delta (G)\le 5$.

Prove five colour theorem by induction on vertex count. Remove a vertex $x$ of degree at most five and colour the remaining planar graph with five colours. If its neighbours use fewer than five colours, assign an unused colour to $x$. Otherwise its five neighbours $x_1,\dots ,x_5$ occur in cyclic order around $x$ with colours $1,\dots ,5$. If no path of colours $1,3$ connects $x_1$ to $x_3$, interchange these colours in the component of $x_1$, freeing colour 1 at $x$.

If such a path exists, a simple path together with $x{x}_1$ and $x{x}_3$ makes a Jordan curve. The cyclic order puts $x_2$ and $x_4$ on opposite sides. A path of colours $2,4$ between them would have to cross the curve, impossible in the plane embedding; it cannot meet the differently coloured path or use the removed vertex. Interchange colours $2,4$ in the component of $x_2$, freeing colour 2 at $x$. The colouring extends in every case. Thus $\chi (G)\le 5$ for finite simple planar graphs. The usual simple-graph convention is needed: loops would invalidate a proper-colouring assertion.

The communicating classes are $T=\{1,2\}$, $B=\{3,4\}$ and $C=\{5,6,7\}$. The class $T$ is open: it can reach $B$ or $C$ and cannot be reached from either. To remain in $T$ for two steps has probability $1/6$, so ${\mathbb{P}}_i(X_{2k}\in T)=(1/6{)}^k$ for $i\in T$. Its states are transient states. All returns within $T$ have even length and a two-step return is possible, so their period of a state in a Markov chain is $2$.

The closed communicating class $B$ alternates deterministically between $3$ and $4$, and therefore consists of positive recurrent states of period $2$. The class $C$ is finite, closed and irreducible, so its states are also positive recurrent states. The self-loop at $7$ gives period $1$ there, and the period is constant on a communicating class, so every state of $C$ has period $1$.

For $C$, solve the stationary distribution equations:Hence $({\pi }_5,{\pi }_6,{\pi }_7)=(1/4,1/4,1/2)$. Finite irreducibility and aperiodicity imply convergence to this stationary distribution from any state of $C$. If $h_i$ is the probability of eventually entering $C$, the Markov property givesso $h_1=2/5$ and $h_2=3/10$. Conditioning on the finite entrance time and then letting the remaining time tend to infinity proves that the limiting probabilities in $C$ are $h_i{\pi }_j$. The probability of entrance after a large cutoff tends to zero, which justifies this conditioning limit.

For $B$, convergence depends on the parity balance on entry to a period-two Markov class. Entry always occurs at $3$. Starting from $1$, the probabilities of first entry at time $2k+1$ and at time $2k+2$ are both $\frac{1}{4}(1/6{)}^k$. Each parity therefore has total probability $3/10$, so both $p_{13}^{(n)}$ and $p_{14}^{(n)}$ tend to $3/10$, despite the recurrent class being periodic. Starting from $2$, the corresponding odd- and even-entry probabilities are $\frac{1}{2}(1/6{)}^k$ and $\frac{1}{12}(1/6{)}^k$, with totals $3/5$ and $1/10$. Consequentlyso neither limit exists. From $3$ or $4$, the probabilities at $3$ and $4$ alternate between zero and one; from $C$ they are identically zero. Finally every probability of being in $T$ tends to zero, from all starting states.

All $49$ requested conclusions are encoded below, with $\ast$ denoting a nonexistent limit:The class classification is $T$: transient, period $2$; $B$: recurrent, period $2$; $C$: recurrent, period $1$.

If the primed frame moves at speed $v$ in the positive $x$ direction relative to the unprimed frame, the one-dimensional Lorentz transformation isFor a trajectory with velocity $q=dx/dt$, differentiating these expressions gives $q^{\prime }=(q-v)/(1-vq/c^2)$. Solving for $q$, when the particle has primed-frame speed $u$, proves the relativistic velocity-addition formulaFor the train problem the sign of the frame velocity reverses: the train frame $S$ moves at $+v$ relative to the station frame $S^{\prime }$. Put $\gamma =(1-v^2/c^2{)}^{-1/2}$. The transformation appropriate to the subsequent parts is therefore

For a parametrized curve on the surface, the energy of a curve iswhere the first fundamental form coefficients are evaluated at $(u(t),v(t))$. A fixed-endpoint variation of a curve has coordinate variation fields $\eta ,\xi$ vanishing at $a,b$. Differentiating under the integral and integrating their derivative terms by parts givesThe fields can be chosen independently with compact support in $(a,b)$. Stationarity and the fundamental lemma of the calculus of variations therefore force both coefficients to vanish. These Euler-Lagrange equations are the geodesic equations:

A topological space is connected if it cannot be written as the union of two disjoint nonempty open subsets. Equivalently its only subsets that are both open and closed are the empty set and the whole space. For a subspace topology, open means relatively open within that subspace. A path-connected space is connected, since a separation would pull back along a path to a separation of an interval.

The determinant isEquivalently it is the alternating multilinear function of the columns with value one on the identity matrix. To prove multiplicativity, write column $j$ of $AB$ as ${\sum }_k{b}_{kj}{A}_k$, where $A_k$ is column $k$ of $A$. Multilinear expansion gives a sum over choices $k_1,\dots ,k_n$; terms with repeated choices vanish by alternation. The remaining permutations contributeIf $C$ is invertible, $CB=C(BC)C^{-1}$, so multiplicativity makes $\det (CB{C}^{-1}-\lambda I)=\det (BC-\lambda I)$ and $p_{CB}=p_{BC}$. For arbitrary $A$, put $C=A+tI$. Its determinant is a nonzero polynomial in $t$, so it is invertible except for finitely many values. Thus for all other $t$,For each fixed $\lambda$ both sides are polynomials in $t$, and equality at infinitely many $t$ makes them identical. Setting $t=0$ proves the Characteristic polynomials of AB and BA identity $\boxed{p_{BA}=p_{AB}}$ without an invertibility assumption on $A$.

In the convention $p_A(\lambda )=\det (A-\lambda I)$, the coefficient of ${\lambda }^{n-1}$ is $(-1{)}^{n-1}\mathrm{tr}A$: in the determinant expansion such a term chooses one diagonal entry of $A$ and $n-1$ factors $-\lambda$; nonidentity permutations have at least two nonfixed positions and contribute lower degree. Therefore $\boxed{p_A=p_B⟹\mathrm{tr}A=\mathrm{tr}B}$.

For $n\ge 1$, an eigenvalue is a number $\lambda \in \mathbb{C}$ for which there is a nonzero vector $v$ satisfying $Av=\lambda v$. Equivalently, $\ker (A-\lambda I)\ne \{0\}$, or $\det (A-\lambda I)=0$. The characteristic polynomial $\det (tI-A)$ has degree $n$ and leading coefficient one. By the fundamental theorem of algebra it has a complex root $\lambda$. At that root $A-\lambda I$ is a singular matrix, so it has a nonzero kernel vector, proving that $A$ has an eigenvalue.

Distinguish the full velocity potentials ${\Phi }_i=U_{i}x+{\phi }_i$ from their perturbations ${\phi }_i$. The linear boundary conditions use the perturbations. This distinction reconciles the notation in the printed conditions with the full potentials written in the mode ansatz. All equations below neglect products of perturbations. Assume positive densities and no surface tension.

A topology on $X$ is a family $\mathcal{T}$ of subsets containing $\varnothing$ and $X$, closed under arbitrary unions and finite intersections. Its members are the open sets. A Hausdorff space separates each pair of distinct points by disjoint open neighbourhoods; a compact space is one for which every open cover has a finite subcover.

Let $\Omega$ be the Earth's angular velocity magnitude. Use the local east, north and upward unit vectors ${\mathbf{e}}_E,{\mathbf{e}}_N,{\mathbf{e}}_U$, with ${\mathbf{e}}_E\times {\mathbf{e}}_N={\mathbf{e}}_U$. At latitude $\lambda$, the Earth's angular velocity and the train's ground-relative velocity areThus the east component of the Coriolis force isFor a north–south track, the train has no eastward acceleration relative to Earth. Its path curvature lies in the local north–vertical plane. The centrifugal acceleration and Newtonian gravity also lie in that plane, and the Earth's constant angular velocity gives no Euler acceleration. Hence the sideways force from the track must cancel the eastward Coriolis force:Its magnitude is $2m\Omega V\sin \lambda$, and its direction is west in the Northern hemisphere. The eastward tendency is the apparent Coriolis force; the requested track force has the opposite direction.

At infinity, the meteorite's incoming asymptote has speed $V$ and perpendicular distance $b$ from the Earth’s centre. The magnitude of its specific angular momentum is the magnitude of $\mathbf{r}\times \mathbf{v}$. On the asymptote, this is $bV$, so conservation of angular momentum gives $\boxed{h=bV}$. Its mechanical energy at infinity is $E=m{V}^2/2$. Comparing with part (ii) givesFor $b>0$ this Kepler orbit is a hyperbola. The smallest $r=1/u$ occurs at periapsis, where $\cos (\theta -{\theta }_0)=1$. ThusThe middle equality follows by substituting $e^2-1=b^2{V}^4/(G^2{M}^2)$. A trajectory entering the Earth satisfies $r_{\min }<R$, equivalentlyTherefore the sufficient collision condition isThis is gravitational focusing: Newtonian gravity increases the collision radius measured at infinity. Equality is the ideal grazing trajectory, which touches the surface; a strict inequality penetrates it. For $b=0$ the incoming motion is radial and collides, with $r_{\min }=0$ as the limiting value of the above expression.

Write a Möbius transformation as $M(z)=(az+b)/(cz+d)$ with $ad-bc\ne 0$, interpreted on the Riemann sphere. Let $T_u(z)=z+u$, $D_{\lambda }(z)=\lambda z$ with $\lambda \ne 0$, and $I(z)=1/z$.

If $c=0$, then $a,d\ne 0$ and $M(z)=(a/d)z+b/d=T_{b/d}\circ D_{a/d}(z)$. If $c\ne 0$, algebraic division givesConsequentlyThe scaling coefficient is nonzero because $ad-bc\ne 0$. Thus translations, nonzero complex scalings and reciprocal inversion generate every Möbius map. The formulas extend over their poles and infinity by the sphere interpretation.

For a normal subgroup $H⊴G$, define multiplication of left cosets byTo check that this is well-defined, replace $g,k$ by $g{h}_1,k{h}_2$ with $h_1,h_2\in H$. Their product is $g{h}_{1}k{h}_2=gk(k^{-1}{h}_{1}k)h_2$, and normality puts $(k^{-1}{h}_{1}k)h_2$ in $H$. Thus the product coset is independent of the representatives. Associativity follows from multiplication in $G$, the identity is $H$, and the inverse of $gH$ is $g^{-1}H$. These operations form the quotient group $G/H$.

For the unheaded finite-index conclusion, let $G$ act on the $n$ left cosets of its proper subgroup $H$ by $g\cdot (xH)=gxH$. This is well-defined even if $H$ is not normal: replacing $x$ by $xh$ leaves $gxH$ unchanged. The maps are bijections with inverses supplied by $g^{-1}$, and their compositions give a group homomorphism $\phi :G\to S_n$. Its kernel of a group homomorphism $N$ is normal, by the calculation in part (ii).

The action is not trivial, because some $g\notin H$ sends the coset $H$ to $gH\ne H$; hence $N\ne G$. If $N$ were trivial, $\phi$ would be injective, implying $|G|\le |S_n|=n!$. Under the given strict inequality this is impossible. Thus $N$ is a nontrivial proper normal subgroup, andMore precisely, $N={\bigcap }_{x\in G}xH{x}^{-1}$ is the subgroup core: $g$ fixes every coset if and only if $x^{-1}gx\in H$ for every $x$. This identifies the normal subgroup produced by the coset action.

Use the right-handed rotation matrix which sends the positive $x$ direction to the positive $y$ direction:The Cartesian second-rank tensor transformation law givesHere isotropic tensor means invariant under every proper rotation, as is needed for the rank-three assertion involving the Levi-Civita symbol. A passive coordinate rotation through the same angle uses $R^{\mathsf{T}}$ instead; the invariance conclusions are identical.

Setting $S^{\prime }=S$ forces $S_{11}=S_{22}$, $S_{12}=-S_{21}$, and $S_{13}=S_{23}=S_{31}=S_{32}=0$. No symmetry assumption on $S$ has been made: it now has the formInvariance under a right-handed quarter-turn about the $x$ axis next gives $S_{22}=S_{33}$ and $S_{12}=-S_{13}$, hence $a=c$ and $b=0$. Conversely $R(\lambda I)R^{\mathsf{T}}=\lambda I$ for every orthogonal matrix. The general isotropic second-rank tensor is therefore $S_{ij}=\lambda {\delta }_{ij}$.

An isotropic vector must be fixed by the half-turn about the $z$ axis, so its first two components vanish. The half-turn about the $x$ axis then forces its third component to vanish. Thus the only isotropic vector is zero. The general isotropic third-rank tensor under proper rotations is $A_{ijk}=\lambda {ϵ}_{ijk}$. If invariance under the whole orthogonal group were required of an ordinary rank-three tensor, inversion $R=-I$ would send $A$ to $-A$, forcing $A=0$ instead.