The official notes, PDF p. 18, handwritten p. 11 explain the imminent maximum expansion, recollapse, remaining time of order the present age, and total lifetime of approximately twice that age.

The official notes, PDF pp. 17–18, handwritten pp. 10–11 give the matter density, curvature and cosmological constant scalings and the requested logarithmic sketch. For the closed model the curvature contribution to the Friedmann equations is negative; the sketched curvature magnitude must be read with that sign restored.

The official notes, PDF pp. 14–16, handwritten pp. 7–9 integrate the exponential scale factor, discuss the sign required for an incoming photon, and obtain the distance–redshift relation. They use $a(t_e)$ for the paper's otherwise undefined $R(t_e)$.

The official notes, PDF pp. 13–14, handwritten pp. 6–7 integrate the radial geodesic equations and establish the vanishing integration constant for a null geodesic.

The official notes, PDF pp. 11–13, handwritten pp. 4–6 derive all four geodesic equations and list the twelve nonzero Christoffel symbols up to symmetry of the lower indices, using time coordinate $t$.

The official notes, PDF p. 11, handwritten p. 4 give the Lagrangian for the geodesic equations.

The official notes, PDF p. 9, handwritten p. 3 compare the two clock ratios and identify the station's rotational contribution. PDF p. 10, handwritten p. 3a, separately clarifies the normalization when coordinate time is the affine parameter.

The official notes, PDF pp. 8–9, handwritten pp. 2–3 calculate the geostationary satellite/station proper time ratio, including the equatorial station's rotation. They explicitly correct the formula printed in the paper; the printed target is not the first-order result of the stated Schwarzschild metric.

The official notes, PDF pp. 7–8, handwritten pp. 1–2 calculate both elapsed proper times and their weak-field ratio for satellite A and the polar station.

The official notes, PDF p. 7, handwritten p. 1 derive the affine choice of coordinate time and the circular-orbit angular velocity from the geodesic equations.

Put $\omega ={\zeta }^5$, $\tau ={\zeta }^3$, $a=1+2\omega$, and $b=1+2(\tau +{\tau }^{-1})$. The primitive cubic and quintic root identities give $a^2=-3$, $b^2=5$. These elements therefore provide the following six generators:

The three quadratic fields are distinct because $-3,5,-15$ are distinct rational square classes. To justify the fifth entry, the two quadratic fields $\mathbb{Q}(a)$ and $\mathbb{Q}(b)$ are distinct, so their compositum has degree four. Moreover $a+b\ne 0$ andThus their sum alone generates the compositum.

The field $\mathbb{Q}(\tau )$ has degree four by the quintic cyclotomic polynomial. Its Galois group is $C_4$ and its unique quadratic subfield is $\mathbb{Q}(b)$, so it cannot contain $a$. It is therefore different from $\mathbb{Q}(a,b)$. The final field has degree four by part (b) and is real; the preceding two quartic fields contain nonreal $\tau$ and $a$, respectively. Thus all three quartics are distinct, and the count in part (i) proves the list exhaustive. This gives the subfields of the fifteenth cyclotomic field without relying on a choice of square-root signs.

Irreducibility of the cyclotomic polynomial gives $[K:\mathbb{Q}]=\phi (15)=8$. The extension is Galois, since $K$ is the splitting field of $T^{15}-1$ in characteristic zero. The injection in part (a) is consequently an isomorphism, and the Chinese remainder theorem givesThe Fundamental theorem of Galois theory identifies proper intermediate fields strictly above $\mathbb{Q}$ with proper nontrivial subgroups. Such subgroups have order two or four. The two-torsion subgroup is $C_2\times C_2$, with three nonidentity elements and hence three order-two subgroups. There are also three order-four subgroups: they are the kernels of the three nonzero group homomorphisms $C_2\times C_4\to C_2$. Distinct such maps have distinct kernels because the target has only one nonidentity element. Therefore

Beyond the outer classical turning point, let $\chi (x)=\sqrt{2m[V(x)-E]}$ and $U(x)={\hslash }^{-1}{\int }_{x_2}^x\chi (y)dy$. The bound state condition removes the growing branch at positive infinity, leavingHere $F$ is an arbitrary common amplitude pending the matching and wavefunction normalization. For the confining potential energy in the native diagram, the decaying branch gives the required exterior square-integrable function.

In the right-hand classically allowed region, put $p(x)=\sqrt{2m[E-V(x)]}$ and $s(x)={\hslash }^{-1}{\int }_{x_1}^{x}p(y)dy$. A convenient form of the general WKB approximation isThe phase origin $\pi /4$ is a choice of basis for the two oscillatory solutions, anticipating the Airy turning-point connection formula. Equivalently, one may use independent $e^{is}$ and $e^{-is}$ branches. The coefficients have not yet been related to those in either forbidden region.

In the central classically forbidden region, define the positive decay momentum $\chi (x)=\sqrt{2m[V(x)-E]}$ and the dimensionless action integral $K(x)={\hslash }^{-1}{\int }_x^{x_1}\chi (y)dy$. The general leading WKB approximation isBoth coefficients are allowed: this interval has finite length, so the exponential that grows towards the barrier centre cannot be discarded by a condition at infinity. The WKB validity condition holds away from the classical turning point $x_1$; a local Airy function solution supplies its continuation there.

At an equilibrium $(v_{\ast },k{v}_{\ast })$, set $p=f^{\prime }(v_{\ast })$. The Jacobian matrix, its trace and its determinant areThe eigenvalues and their discriminant areThus the complete local linear stability test isIn the last two cases, $\mathcal{D}>0$ gives a node, while $\mathcal{D}<0$ gives a focus; the sign of $T$ distinguishes inward and outward trajectories. At $\mathcal{D}=0$, a nonzero repeated eigenvalue gives a degenerate node. At $p=\gamma <k$ the eigenvalues are purely imaginary, so linear stability analysis alone does not decide asymptotic stability; with the required crossing and nondegeneracy this is a Hopf bifurcation. At $p=k$ there is a zero eigenvalue, requiring a nonlinear test and generically marking the saddle-node bifurcation in the preceding answer.

For a unique simple equilibrium, $p<k$, so it can be a sink equilibrium or a source equilibrium. The number of equilibria alone does not decide stability. If $k\le \gamma$, the unique simple equilibrium is always a sink equilibrium. If $\gamma <k$, it can become a source equilibrium where $\gamma <p<k$. The same conclusion follows particularly simply if $\gamma >M$: the trace is negative at every nonsaddle equilibrium.

For three simple equilibria ordered by their $v$ coordinates, the middle one has $p>k$ and is a saddle equilibrium; the two outer ones have $p<k$. When both outer equilibria are sink equilibria, each has an attracting neighborhood, giving bistability. In the usual two-basin portrait, the stable manifold of the saddle equilibrium separates their basins. Each branch of its unstable manifold initially leaves towards a different side; its eventual destination requires a global argument. If $k\le \gamma$, both outer equilibria must be sink equilibria. If $\gamma <k$, either or both outer equilibria can instead be source equilibria; the same test $p\lessgtr \gamma$ applies separately to each. The three lower panels illustrate all three possibilities up to exchanging left and right.

The arrows in the phase plane follow directly from the nullclines: below $w=f(v)+I$ the horizontal component points right, and above it left; below $w=kv$ the vertical component points up, and above it down. Near a stable focus trajectories spiral inward; near an unstable focus they spiral outward. Near a saddle equilibrium they approach along its stable manifold and leave along its unstable manifold. These directions, together with the local tests, determine the sketches.

There is also a useful global conclusion when the unique equilibrium is a source equilibrium. Withthe leading term in $v$ is $-\beta v^4$, so sufficiently large ellipses are trapping regions. A small repelling neighborhood of the hyperbolic source equilibrium can be removed to form an invariant annulus. The Poincaré-Bendixson theorem then guarantees at least one periodic orbit in that annulus. The numerical sketch illustrates the resulting sustained oscillation; this argument does not assert uniqueness of a limit cycle. A stable equilibrium can instead represent a resting membrane potential, with perturbations producing damped relaxation; near a threshold, a sufficiently large perturbation can produce a large excursion before recovery. Global basin boundaries and coexistence of periodic orbits need more information than the number and local stability of equilibria.

Put $k=\beta /\gamma$. The nullclines of the space-clamped FitzHugh-Nagumo model areThe first nullcline is a downward-leading cubic polynomial, shifted vertically by $I$; the second nullcline is a line through the origin with slope $k>0$. Their intersections satisfyWriting $M=(1-\alpha +{\alpha }^2)/3$, differentiation givesIf $k>M$, $h$ increases strictly, so there is exactly one equilibrium. If $k=M$, $h$ is still strictly increasing, although its derivative vanishes at one point, so the equilibrium is again unique. If $0<k<M$, defineHere $h$ has a local maximum at $u_{-}$ and a local minimum at $u_{+}$; thereforeAt either endpoint there are two distinct equilibria, one a double root. These are the saddle-node bifurcation boundaries omitted by the generic one-or-three wording. At $k=M$ and $I=h((1+\alpha )/3)$ there is one triple root. This is the Cubic FitzHugh-Nagumo nullcline criterion.

The zero extension of H01 makes $v_m$ an element of $H^1({\mathbb{R}}^3)$. The chain rule and change of variables give the exact scaling identitiesIndeed $\nabla v_m(x)=r_m^{3/2}\nabla u_m(r_{m}x)$ inside the expanded ball, and there is no boundary jump because the Sobolev trace is zero.

First take the printed convergence assertion as an assumption, as the question requests. It would imply $\Vert \nabla v_m{\Vert }_2\to \Vert \nabla v{\Vert }_2>0$. The strict inequality follows because an $H^1({\mathbb{R}}^3)$ function with zero weak gradient is constant almost everywhere, and the only constant in $L^2({\mathbb{R}}^3)$ is zero; this contradicts the asserted positivity of $v$. If $u_m$ converged strongly in $H^1(B)$, its limit would have to be its weak limit zero from (iii), forcing $\Vert \nabla u_m{\Vert }_2\to 0$. The first scaling identity contradicts this. Thus the intended answer is

There is nevertheless a substantive printing error: the asserted strong convergence of $v_m$ in the full $H^1({\mathbb{R}}^3)$ is false for these positive sources. Here is a check that does not rely on any assertion about the limiting profile. Put $R_m=r_m^{-1}$ and $c=|E{|}^{-5/6}$. Rescaling the weak formulation givesThe spherical mean of a Poisson solution outside its source provides an exact obstruction. The total source is $M=c|E|=|E{|}^{1/6}>0$. Let $a_m(s)=(4\pi {)}^{-1}{\int }_{S^2}{v}_m(s\omega )d\omega$ be its spherical mean. On $1<s<R_m$ the function is a harmonic function. The radial flux identity, obtained by testing against radial test functions or by the divergence theorem, gives $-4\pi s^2{a}_m^{\prime }(s)=M$, while the zero boundary value gives $a_m(R_m)=0$. HenceCauchy-Schwarz inequality on each sphere, followed by spherical coordinates, now yieldsThus the printed sequence is not even bounded in the full $H^1$ norm. The result on Poisson convergence on expanding balls in a homogeneous Sobolev space gives the appropriate corrected convergence in the homogeneous Sobolev space ${\dot{H}}^1({\mathbb{R}}^3)$, which controls the gradient in $L^2$ and the function in $L^6$, without requiring it to belong to $L^2$.

For completeness, the desired nonconvergence of $u_m$ has a direct proof that needs neither the incorrect claim nor its correction. Choose a nonnegative nonzero test function $\eta \in C_c^{\infty }(E)$, which exists since $E$ is a domain containing the origin. The rescaled weak formulation and Cauchy-Schwarz inequality giveConsequently, for every $m$,Together with (iii), this rigorously proves failure of strong convergence for the actual sequence defined in the paper.

Take the weakly convergent subsequence in (i), with limit $u\in H_0^1(B)$. For each $\varphi \in H_0^1(B)$, the weak formulation givesThe left side tends to $\int \nabla u\cdot \nabla \varphi$ by weak convergence in $H^1$. The zero-boundary Sobolev inequality gives $\varphi \in L^6(B)$, so (ii) makes the right side tend to zero. Taking $\varphi =u$ in the limiting identity yields ${\int }_B|\nabla u{|}^2=0$. The zero-boundary Poincaré inequality then gives $u=0$.

The same argument applies to a weakly convergent subsequence of any subsequence of $(u_m)$, because all remain bounded as in (i). To see why this proves convergence of the full sequence, suppose a continuous linear functional $F$ on $H^1(B)$ did not satisfy $F(u_m)\to 0$. Some subsequence would have $|F(u_{m_j})|\ge \epsilon >0$. Its boundedness gives a further weakly convergent subsequence, whose limit has just been proved to be zero; then $F$ tends to zero along it, a contradiction. ThereforeUniqueness of a possible limit alone is not enough without the weak sequential compactness in a Hilbert space used to extract a further subsequence.

The Lp norm of the forcing at the critical exponent is constant:To prove weak convergence in $L^{6/5}(B)$, the duality of Lp spaces says that we must test against an arbitrary $g\in L^6(B)$. Holder inequality, restricted to the shrinking measurable set $E_m$, givesThe last step is absolute continuity of the Lebesgue integral, since $|E_m|=r_m^3|E|\to 0$ and $|g{|}^6$ is integrable. ThusTesting only bounded or smooth functions would not by itself establish this weak convergence; the restricted Holder inequality handles every member of the full dual space.