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On the unit circle the two equations reduce to the same conditionsince $y^2=1-x^2$. Define $H(x,y)=x^2(f(x,y{)}^2+g(x,y{)}^2)-f(x,y{)}^2$. At either point $(\pm 1,0)$ it is strictly positive, and at either point $(0,\pm 1)$ it is strictly negative. On each of the four closed quarter-circle arcs, continuity and the intermediate value theorem give a zero in its interior. These four interiors are disjoint, so there are at least four distinct solutions, one in each open quadrant. No solution here lies on an axis because $f,g$ are strictly positive.

Let $v_I,v_J,v_K$ be the vertices opposite the corresponding sides, and let ${\lambda }_I,{\lambda }_J,{\lambda }_K$ be their barycentric coordinates. We prove the equivalence by negating each assertion.

Suppose the closed covering sets have empty triple intersection. For $x\in T$ put $d_I(x)=\mathrm{dist}(x,A)$, and define $d_J,d_K$ similarly. At least one distance is positive, since $x$ cannot belong to all three sets, and at least one is zero, since they cover $T$. Consequentlyis continuous and lies in the boundary: one barycentric coordinate vanishes. On $I$ its first coefficient is zero because $I\subseteq A$, so $f(I)\subseteq I$, and likewise for the other sides. This contradicts assertion (1), proving (1) implies (2).

Conversely, suppose such a boundary-valued map exists. Define closed subsets of ${\mathbb{R}}^2$ byThey are closed because $T$ is compact and the coordinate functions are continuous. They cover $T$ because $f$ is boundary-valued; the side-preserving condition gives $I\subseteq A$, $J\subseteq B$, $K\subseteq C$. Their triple intersection is empty because barycentric coordinates sum to one. Thus failure of (1) implies failure of (2), completing the equivalence.

In fact both assertions hold. If the map in (1) existed, let $R$ rotate the equilateral triangle through $120^{\circ }$. Brouwer fixed-point theorem applied to $R\circ f:T\to T$ gives a fixed point $x$ on the boundary. On any side $I$ containing $x$, side preservation puts $x$ also on $R(I)$, so it is their common vertex. A side-preserving map fixes every vertex because that vertex lies on two sides. Therefore $x=Rf(x)=Rx$, impossible for a noncentral vertex. This also proves the covering-intersection assertion through the established equivalence.

The planar Brouwer fixed-point theorem states that every continuous map from a nonempty compact convex subset of ${\mathbb{R}}^2$ to itself has a fixed point. In particular it applies to a closed disk or a closed triangle. Convexity and compactness are essential hypotheses in this formulation.

For a relativistic species at zero chemical potential, $p=ϵ/3$ and $s=(ϵ+p)/T_i$, so its entropy density is proportional to $g_i{T}_i^3$. The Bose-Einstein distribution gives the bosonic normalization, while the fermionic thermal integral is $7/8$ of it. Species still in thermal equilibrium have $T_i=T$; freely streaming decoupled species can have another temperature and therefore contribute an extra factor $(T_i/T{)}^3$ when the total is expressed in units of $T^3$. This gives the two terms $N_{\ast }$ and $N_{SD}$ and their stated effective-degree-of-freedom weights.

After neutrino decoupling, the comoving entropy of the photon-electron-positron plasma is conserved separately from the neutrinos. Before electron-positron annihilation its effective entropy count is $2+(7/8)4=11/2$; afterwards it is $2$. Consequentlywhereas $a{T}_{\nu }$ stays equal to its pre-annihilation value. HenceThis is the instantaneous-decoupling, negligible residual-electron approximation used in the question. Heating of the still-coupled plasma, not cooling peculiar to neutrinos, creates the temperature difference.

Substitute $x=pc/(k_{B}T)$ in the Fermi-Dirac distribution energy integral. ThenThe integral converges, is positive, and is independent of temperature. Thus $ϵ=\alpha T^4$, with $\alpha$ equal to the displayed constant integral. The given normalization is for the specified degeneracy; further internal degrees of freedom multiply it.

The procedure in part (i) sometimes needs five tests, for items one or two. A binary decision tree of height $N$ has at most $2^N$ leaves, so distinguishing ten possibilities requires $N\ge \lceil {\log }_{2}10\rceil =4$.

A prefix code attaining this worst-case bound isUse the same subset-testing rule. The minimum worst-case number is four, so the expected-optimal procedure is not worst-case optimal. This four-test procedure has expectation $175/55$, illustrating the small cost of imposing the smaller maximum depth.

There is a genuinely different optimal procedure. Exchange the codewords assigned to items three and four. Both have code length four, so each item's length, and hence the expected number of tests $173/55$, remains unchanged. The first tested subset now contains item three instead of item four, so this changes the tested groups rather than merely reversing the names of the two test outcomes. The new prefix code still identifies every item uniquely and retains the minimum Huffman coding expectation.

Each test divides the current candidate set into a tested subset and its complement: failure places the defective item inside the subset, whereas success places it outside. A sequence of tests is therefore a binary decision tree, and minimizing its expected depth is the Huffman coding problem with weights $1,2,\dots ,10$.

Combine the smallest two current weights repeatedly. One choice of merges is $1+2=3$, $3+3=6$, $4+5=9$, $6+6=12$, $7+8=15$, $9+9=18$, $10+12=22$, $15+18=33$, $22+33=55$. It gives this explicit prefix code:At any node, test the candidate items whose next code digit is one. Assign failure to digit one and success to digit zero, and repeat until only a single codeword remains. This specifies every test, including the first subset $\{4,5,7,8,9\}$. Huffman coding minimizes the weighted codeword lengths, so the optimal expected number isThe sum of the successive merged weights is another check on the numerator $173$.

Let $k_0=\lceil (n+1)/2\rceil$, $\theta =\pi /(n+1)$, and define the high-frequency endpoint parametersThey obey $1\le a\le b<2$ (with equality of endpoints for a single high mode). Since $|1-\omega t|$ is convex in $t$ and the endpoint modes are included, the exact high-frequency attenuation isThe minimax choice balances the endpoint values with opposite signs. If they were not balanced, moving $\omega$ toward their balancing point would lower the larger endpoint value. ThusFor odd $n$, $a=1$, giving ${\omega }_{\ast }=2/(2+\cos \theta )$ and ${\mu }_{\ast }=\cos \theta /(2+\cos \theta )$. For even $n$, $a=1+\sin (\theta /2)$, givingThese give the exact finite-grid high-frequency Jacobi smoothing optimum, including complete annihilation when there is only one high-frequency mode. Their grid-independent limits are $\boxed{{\omega }_{\ast }=2/3,{\mu }_{\ast }=1/3}$. Choosing $2/3$ damps every high mode by at most $1/3$ on every grid, while its low-mode factor remains $1-O(n^{-2})$. Thus high frequencies decay geometrically much faster. The familiar $2/3,1/3$ pair is the optimal uniform-in-grid smoother; it need not be the exact optimum on a particular finite sine-mode grid.

For a fixed $n$, convergence for every initial error is equivalent to $|1-\omega t_k|<1$ for all $k$, where $t_k=1-\cos (k\pi /(n+1))>0$. ThusA single parameter works for every finite $n$ precisely when $\boxed{0<\omega \le 1}$: the largest $t_k$ tends to two as $n\to \infty$, so every $\omega >1$ eventually fails, whereas $\omega =1$ has all eigenvalues strictly between $-1$ and one for each finite grid.

For such $\omega$ the lowest-frequency eigenvalue satisfies, using $1-\cos x\le x^2/2$,An initial error proportional to $v_1$ retains exactly the factor ${\lambda }_1^{\nu }$. Hence the worst-case spectral radius cannot improve on a factor of the form $(1-c/n^2{)}^{\nu }$, with $c$ independent of grid size, and reducing a slow-mode error by a fixed fraction takes order $n^2$ iterations. This is a worst-case rate assertion, not a lower bound for every individual initial error: a vector containing only faster modes, or the zero vector, need not exhibit the slow mode.

Subtracting the fixed-point equation for the exact solution givesHere $D=2I$, so the supplied sine eigenvectors of $A$ are also eigenvectors of $H_{\omega }$. Explicitly,Their distinct eigenvalues for nonzero $\omega$ make them an eigenbasis; alternatively the discrete sine orthogonality gives a basis even when $\omega =0$.

Assume the diagonal entries of $A$ are nonzero. The Relaxed Jacobi method simultaneously updates every component using the preceding iterate:In matrix form, with $D=\mathrm{diag}(A)$, this is $x^{(\nu +1)}=x^{(\nu )}+\omega D^{-1}(b-A{x}^{(\nu )})$. The ordinary Jacobi method has $\omega =1$.

Vertical hydrostatic balance with atmospheric pressure at $y=h(x)$ gives $p(x,y)=p_0+\rho g(h(x)-y)$ to this order. Therefore $p_x=\rho g{h}^{\prime }$ and integrating the gradient from part (i), with $h\to 0$ on both sides, givesFor positive $U$ the upstream surface rises and the downstream surface falls. Thus the displacement scale is $\mu |U|\sqrt{a}/(\rho g{b}^{3/2})$, and the flat-geometry consistency condition isIndeed ${\max }_s|s/(1+s^2{)}^2|=9/(16\sqrt{3})$, so this scale controls the maximum, not only a typical value. Together with $b/a\ll 1$ and $\rho |\Omega |\sqrt{a}{b}^{3/2}/\mu \ll 1$, these conditions ensure a thin, weakly displaced, viscosity-dominated gap. The slope is smaller still, of order $(|h|/b)\sqrt{b/a}$.

Here $\ell \sim \sqrt{ab}$, so the geometric condition is $b/a\ll 1$. Let $U$ be the signed roller velocity near its top, with magnitude $a|\Omega |$; choose the rotation sense making $U=a\Omega$ positive in the $x$ direction. The small-inertia condition iswith $|U|$ understood for a reversed rotation. The roller's slope and change of tangential speed are small in the gap region. The liquid is Newtonian, the roller imposes no slip and the free surface is shear-free; surface tension is neglected as in the gravity-only model.

For the leading flat film geometry, use $y=-d(x)$ as the roller boundary and $y=0$ as the free surface. The lubrication approximation gives $\mu u_{yy}=p_x$, $u(-d)=U$, $u_y(0)=0$, henceThe corresponding vertical velocity is determined by incompressibility, for example $v(x,y)={\int }_y^0{u}_x(x,s)ds$, so that $v(0)=0$ and $v(-d)=-U{d}^{\prime }$ by differentiating the constant-flux equation. This is the leading tangency condition at the roller. Steadiness makes $Q$ independent of $x$.

For the rotating roller under a nearly flat free surface, the excess pressure tends to zero in the outer fluid on both sides of the gap. ConsequentlySet $x=\sqrt{2ab}s$. The supplied recurrence gives $I_2=\pi /2$ and $I_3=3\pi /8$, soThe extension of the parabolic gap integrals to infinity is the usual matched thin-gap approximation; their dominant contribution comes from $x=O(\sqrt{ab})\ll a$. Taking the surface flat refers to this leading geometry, not setting its small hydrostatic displacement identically to zero in the pressure balance. An exactly flat stress-free surface at exactly constant pressure could not sustain this variable pressure gradient; the displacement needed to do so is checked in part (ii).

Use signature $(-,+,+,+)$, for which $\square =-{\partial }_t^2+\Delta$. With the Lorenz constraint in Proca theory, the Proca equation is $(\square -m^2)A^a=-{\mu }_0{J}^a$. A static solution therefore satisfies $(\Delta -m^2)A^a=-{\mu }_0{J}^a$.

For $R>0$ direct radial differentiation gives $(\Delta -m^2)(e^{-mR}/R)=0$. At the origin its $1/R$ singularity contributes $-4\pi {\delta }_0$: integrating the Laplacian over a small sphere gives limiting outward flux $-4\pi$, while the mass term's integral tends to zero. Thus $G_m(R)=e^{-mR}/(4\pi R)$ obeys $(\Delta -m^2)G_m=-{\delta }_0$. Convolution proves the printed Yukawa potential solutionThe integral requires suitable source decay, for example compact support. For conserved static sources, differentiating the convolution and integrating by parts gives ${\partial }_a{A}^a=0$ as well; its spatial divergence is the convolution with ${\partial }_i{J}^i=0$.

For the usual nonnegative mass parameter this is the decaying static solution, and $m=0$ gives the Coulomb kernel with Lorenz gauge imposed. If the arbitrary real parameter is negative, the printed kernel still gives a particular solution for compact sources, but grows at infinity; the decaying boundary condition selects $e^{-|m|R}/(4\pi R)$ instead. Static sources also allow homogeneous time-dependent solutions, so this construction establishes a static solution, not that every solution is static.

Take ${\partial }_b$ of the Proca equation. Commuting partial derivatives and antisymmetry make ${\partial }_b{\partial }_a{F}^{ab}=0$, while current conservation gives ${\partial }_b{J}^b=0$. Thus $m^2{\partial }_b{A}^b=0$. For $m\ne 0$, $\boxed{{\partial }_b{A}^b=0}$. This Lorenz constraint in Proca theory is a consequence of the massive equations, rather than a gauge choice.

The stated “arbitrary constant” includes a genuine exception at $m=0$. Then the divergence condition is not implied: with $J=0$ and $A_a={\partial }_a\chi$, $F=0$ solves the equations for any smooth $\chi$, whereas ${\partial }_a{A}^a=\square \chi$ need not vanish. In the massless theory the Lorenz condition can instead be imposed as a gauge fixing.

For compactly supported variations, antisymmetry of the electromagnetic field tensor gives $\delta (-F_{ab}{F}^{ab}/4)=-F^{ab}{\partial }_a\delta A_b$. Integrate by parts in the action to obtainArbitrariness of the variation gives the Proca equationtogether with the unchanged identity ${\partial }_{[a}{F}_{bc]}=0$ coming from $F=dA$. The signs follow from the source and mass terms as printed.

For the quadratic potential the characteristic curves are harmonic oscillations:The inverse is $x_0=x\cos t-v\sin t$, $v_0=v\cos t+x\sin t$. Therefore the Liouville equation has solutionThis formula works for smooth data or by transport for locally integrable data, and for measures by pushforward. Although the quadratic $V$ itself is not globally Lipschitz as in the initial sufficient hypotheses, its gradient is globally Lipschitz, which is the property actually needed for this global flow.

Formally the Jacobian determinant of the characteristic flow map satisfies $J^{\prime }=\mathrm{div}B({\Phi }_t)J=0$, $J(0)=1$, so $J=1$. Changing variables $z={\Phi }_t(z_0)$ givesTaking the $p$th root proves $\boxed{\Vert f(t){\Vert }_{L^p}=\Vert f_I{\Vert }_{L^p}}$ for $1\le p<\infty$. Equivalently multiply the equation by $p{f}^{p-1}$ and integrate the divergence term, with vanishing boundary terms. Volume preservation gives the result without needing such boundary decay for general integrable data.

Along a characteristic curve, the Liouville equation gives $d[f({\Phi }_t(z_0),t)]/dt=0$. HenceComposition preserves nonnegativity, so $f\ge 0$. For locally integrable data this formula defines the transported weak solution, and the Lipschitz invertible flow preserves null sets locally. Thus the same assertion holds almost everywhere without requiring pointwise smooth data.