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Maximize the Gibbs entropy subject to normalization and the two mean constraints. Introduce multipliers $\eta ,A,B$ and differentiateStationarity gives ${\rho }_i\propto \exp [-(A{E}_i+B{N}_i)/k]$. Strict concavity of entropy gives the constrained maximum, on its support. Differentiating the maximum value with respect to the constrained means, or differentiating the entropy along the maximizing distribution and using ${\sum }_{i}d{\rho }_i=0$, gives $dS=AdE+BdN$. The thermodynamic identification is $A=1/T$, $B=-\mu /T$. Therefore

This is endpoint asymptotics after odd Gaussian cancellation. All interior algebraic terms cancel because $\sin t{e}^{-\lambda t^2}$ is odd. Split the interval into $[-1,1]$ and the remaining tail:The right-hand integral has its minimum phase at its left endpoint, with ${\varphi }^{\prime }(1)=2$ and nonzero amplitude $\sin 1$. Endpoint Laplace method therefore giveThe answer is exponentially small, rather than zero; applying only the interior Gaussian approximation misses the asymmetric boundary tail.

The unique minimum is at zero with phase $t^2$ and amplitude $\cos 0=1$. The interior Laplace method givesFor an additional check, extending to the real line gives the exact Gaussian Fourier integral $\sqrt{\pi /\lambda }{e}^{-1/(4\lambda )}$. The omitted tails are exponentially small, so the displayed leading term follows directly.

Suppose first that the unique interior minimum is nondegenerate, ${\varphi }^{\prime \prime }(c)>0$, and $f(c)\ne 0$. Outside a small neighbourhood of $c$ the phase exceeds $\varphi (c)$ by a fixed positive amount, so that region is exponentially smaller. Inside, expandThe rescaling $s=\sqrt{\lambda {\varphi }^{\prime \prime }(c)}(t-c)$ and a dominated local Gaussian calculation give the Laplace methodIf the minimum is at $a$ and ${\varphi }^{\prime }(a)>0$, the scale is $t-a=O({\lambda }^{-1})$, giving $f(a)e^{-\lambda \varphi (a)}/[\lambda {\varphi }^{\prime }(a)]$. At $b$ with ${\varphi }^{\prime }(b)<0$, replace the denominator by $\lambda |{\varphi }^{\prime }(b)|$. A stationary quadratic endpoint gives half the interior Gaussian factor.

The printed assumptions do not guarantee nondegeneracy or $f(c)\ne 0$. More generally, if $\varphi (t)-\varphi (c)\sim A|t-c{|}^m$ with $A>0$ and even $m$ at an interior minimum, the leading phase scale is ${\lambda }^{-1/m}$ and, for $f(c)\ne 0$, the factor isAt a one-sided endpoint the factor two is absent and the first positive one-sided phase power need not be even. If the amplitude vanishes, expand it to its first contributing term; parity can cancel interior terms. A phase flat to all orders requires its actual local behavior rather than a nonexistent quadratic Taylor coefficient.

For the interior stationary candidate, let $\gamma =1/(\alpha +\beta -1)$. Its formal relation $\ell =1-\beta c/\alpha$ givesIntegration therefore gives $x=A{e}^{rt}+B{e}^{-\gamma rt}-1/r$. Imposing $x(0)=x_0$ and $x(T)=0$ yieldsThis recovers the printed expression algebraically, but does not turn its stationary control into an optimum; it may also violate $\ell \ge 0$.

The actual globally optimal wealth from part (ii) isThe budget equation makes the two branches agree at the switch. Direct differentiation gives $\dot{x}=rx+\ell -c$ on both branches, and $x(T)=0$. This solution remains nonnegative throughout, even though only the terminal nonnegativity constraint was imposed. It provides the complete corrected solution for $\alpha +\beta >1$.

The adjoint equation is $\dot{p}=-rp$, so $p(t)=p_0{e}^{-rt}$. For the interior stationary candidate from part (i), writing $s=\beta c/\alpha$ gives $p\propto c^{\alpha +\beta -1}$. Therefore it formally haswhich derives the printed formula as a stationary, generally nonoptimal candidate.

For the work-then-consume optimum for a nonconcave consumption-leisure utility, the endpoint maximization in part (i) giveswhenever consumption is chosen, and $\ell =1,c=0$ otherwise. Compare the two endpoint values: leisure and consumption are preferred whenBecause the unconstrained consumption is increasing in time, there is at most one switch: work first, then cease work and consume at an increasing rate.

Here is a global verification, not just a Pontryagin maximum principle condition. The terminal constraint is equivalent toFor any $p_0>0$, pointwise maximization of $c^{\alpha }{s}^{\beta }-p_0{e}^{-rt}(c+s-1)$ gives an upper bound on every feasible objective, after adding $p_0{x}_0$. A control attaining this pointwise maximum and using the full budget attains the bound, so it is globally optimal.

Let $a=r\alpha /(1-\alpha )$. Ifthe optimal switch time is $t_{\ast }=0$ and $C=C_0$. Otherwise choose the unique $t_{\ast }\in (0,T)$ such thatThe left side of the second equality after substitution decreases strictly from $c_{\ast }(e^{aT}-1)/a$ to zero, while the right side increases from $x_0$ to a positive number. Thus the switch exists and is unique. The control is $(\ell ,c)=(1,0)$ before $t_{\ast }$ and $(0,C{e}^{rt/(1-\alpha )})$ after it. It saturates the budget and attains the dual upper bound, completing the proof of optimality for the actual printed parameter range.

The requested properties are not properties of the optimum under the printed assumption $\alpha +\beta >1$. We first derive their origin, and then solve the actual constrained problem.

Put $s=1-\ell$, so $0\le s\le 1$, and write the Hamiltonian of an optimal-control problemFor an interior stationary control, $H_c=H_s=0$ givesThis proves the printed identity only as a necessary stationarity condition for an interior candidate.

It fails the required maximization condition. For fixed positive $p$ and fixed $s$, maximization over $c$ givesSince $\beta /(1-\alpha )>1$, this is strictly convex in $s$. Its maximum on $[0,1]$ is at an endpoint, not at its interior stationary point. Thus the actual optimum uses $\ell =1,c=0$, or $\ell =0,c>0$. On any consumption interval the printed relation would require $\beta c=\alpha$, although the actual optimal consumption varies with time.

This proves a genuine failure of the printed premise, not just failure to check a boundary condition. If the parameter inequality were reversed and an interior solution stayed within the control bounds, these first-order equations would instead arise from a concave objective.

When $\mu =\lambda$, an excursion after the jump from zero to one visits $n$ before its reset with probabilityThis tends to zero, so a catastrophe eventually occurs with probability one. Both the original chain and its jump chain are recurrent.

The expected number of jumps in a return cycle isThus the jump chain is null recurrent. But its holding time at $n\ge 1$ has mean $1/[\lambda (n+2)]$, and the initial holding time at zero has mean $1/\lambda$. The expected elapsed cycle time isThe continuous-time Markov chain is therefore positive recurrent. Its equilibrium distribution, computed as mean time in a state per mean cycle time, isThe entries sum to one by telescoping. The contrast is explained by progressively shorter holding times at the large states, not by a change in return probabilities.

For this linear birth process with catastrophes, the off-diagonal entries of the Q-matrix arewith all other off-diagonal entries zero. The diagonal entries are $q_{00}=-\lambda$ and $q_{nn}=-[\lambda (n+1)+\mu ]$ for $n\ge 1$. At zero there is no separate catastrophe transition, because it would not change the state.

The chain is irreducible: every positive state can jump to zero, and from zero any finite sequence of births has positive probability. To prove nonexplosion, couple it with the linear birth process with immigration having birth rate $\lambda (n+1)$ and no catastrophes. Its mean population is finite at each finite time, obtained from $m^{\prime }=\lambda (m+1)$, or its exponential holding-time sum diverges since ${\sum }_{n}1/[\lambda (n+1)]=\infty$. Couple births monotonically and implement catastrophes with an independent rate-$\mu$ Poisson clock, resetting the smaller population. On a finite time interval the dominating pure-birth process has finitely many births, and the catastrophe clock finitely many rings. Thus the original chain is nonexplosive.

For its jump chain, the transitions are

The literal assertion needs $q\ne p$. If $q=p$, the two geodesic circles coincide, have equal tangent lines at every point, and do not intersect transversally; their intersection is not finite. We prove that close geodesic circles intersect twice for nearby distinct centres, and in fact find exactly two intersections.

Choose a strongly convex normal neighbourhood about $p$ and a sufficiently small $r>0$ so that every circle considered and all geodesics joining its points to nearby centres remain there. Squared distance $d(q,x{)}^2$ is then smooth in both arguments, and the exponential map is nonsingular on the relevant radius-$r$ tangent circles. Consequently all these distance spheres are smooth embedded one-dimensional manifolds.

Write $q={\exp }_p(\epsilon v)$ with $|v|=1$ and $\epsilon >0$, and parameterize $S_r(p)$ by $x(\theta )={\exp }_p(re(\theta ))$. DefineThe first variation of geodesic energy givesIndeed differentiating the energy of the joining geodesic cancels the interior term by the geodesic equation, leaving the initial endpoint term $-2⟨re(\theta ),v⟩$. This is also the endpoint form of the Gauss lemma.

Since $F(0,v,\theta )=0$, the quotient $F/\epsilon$ extends smoothly to $\epsilon =0$ by integrating ${\partial }_{\epsilon }F$ from zero to $\epsilon$. At zero it is $-2r\cos (\theta -\arg v)$, with exactly two simple zeros. The implicit function theorem gives two nearby simple zeros for small positive $\epsilon$. Compactness of the unit vector circle makes the choice of small $\epsilon$ uniform in $v$; away from fixed neighbourhoods of those two zeros the cosine is bounded away from zero, so there are no additional zeros.

A nonzero angular derivative at either zero says that the defining function of $S_r(q)$ has a nonzero derivative along $S_r(p)$. Thus their tangent lines differ: the intersections are transverse. Shrinking the neighbourhood of centres accordingly proves

The Gauss lemma says that the radial and angular directions of the exponential map are orthogonal, with radial lengths preserved. On a surface it givesso the metric in geodesic polar coordinates is $d{r}^2+G(r,\theta )d{\theta }^2$.

To prove it, consider a smooth variation $\gamma (r,s)={\exp }_p(rv(s))$ of geodesics with unit initial vectors $v(s)$. Put $T={\gamma }_r$ and $J={\gamma }_s$. Each radial geodesic has constant unit speed, and the Levi-Civita connection is torsion free. HenceAt $r=0$, $J=0$ because all geodesics start at $p$, so the constant inner product is zero. Unit radial length follows from constant speed. General initial-vector variations split into a radial component, whose image has the same length, and a component tangent to the initial sphere, whose image is orthogonal to it. This proves the full Gauss lemma.

For $v\in T_{p}S$ for which the geodesic exists up to time one, the exponential map is ${\exp }_p(v)={\gamma }_v(1)$, where ${\gamma }_v(0)=p$ and ${\dot{\gamma }}_v(0)=v$. The initial velocity determines the geodesic uniquely. Since $d({\exp }_p{)}_0$ is the identity, this map is a local diffeomorphism near zero.

Choose an orthonormal basis of $T_{p}S$ and set $e(\theta )=(\cos \theta ,\sin \theta )$. The geodesic polar coordinates arein a normal neighbourhood, away from its centre. For sufficiently small positive $r$, the geodesic circle iswithin that neighbourhood. Outside the injectivity range the distance sphere need not have this smooth coordinate description.

The same pole-order argument provides a basisFor $n\ge 0$ the first set contains $\lfloor n/2\rfloor +1$ elements, and the second contains $\max (0,\lfloor (n-3)/2\rfloor +1)$. Distinct pole orders imply linear independence, and the coordinate-ring expression proves spanning. HenceFor $n<0$ a function in the space would have no poles anywhere and would vanish at infinity; a regular function on a projective integral curve is constant, so it must be zero. Equivalently the coordinate-ring pole argument first makes it constant, then the vanishing condition excludes nonzero constants.

A rational function with no poles away from $P_{\infty }$ is regular on the smooth affine curve. Its ring of regular functions is the coordinate ringso it has a unique expression $A(x)+yB(x)$ with polynomials $A,B$. This regular-function assertion uses smoothness, hence normality, of the affine curve, not Riemann-Roch theorem.

The nonconstant leading term of $A$ has even pole order $2\mathrm{deg}A$ at infinity, while $yB$ has odd pole order $3+2\mathrm{deg}B$. These different parities prevent cancellation. Thus no nonconstant such function has a pole of order at most one. Therefore

Let $P_j=({\lambda }_j,0)$, and let $Q_{\pm }=(0,\pm \sqrt{-{\lambda }_1{\lambda }_2{\lambda }_3})$. The latter are distinct because the product is nonzero and the characteristic is not two. At each $Q_{\pm }$, the derivative of the curve equation with respect to $y$ is nonzero, so $x$ is a local parameter and has a simple zero. At $P_j$, the derivative of the cubic in $x$ is nonzero, so $y$ is a local parameter with a simple zero; $x-{\lambda }_j$ has order two there.

The only poles occur at the point at infinity, and their orders were computed in part (i). Thus the principal divisors areOver a field where the $Q_{\pm }$ are not rational, their sum is interpreted as the degree-two zero divisor, or computed after scalar extension and then descended.

Use the usual hypothesis that the ground field has characteristic different from two, and work over its algebraic closure for the geometric points. This hypothesis is necessary for the smooth elliptic-curve calculation: in characteristic two, the displayed cubic model can be singular even when its three roots are distinct.

Homogenization givesAt $Z=0$ this says $X^3=0$, soTo compute the pole orders on a smooth cubic in Weierstrass form, work in the chart $Y=1$, writing $u=X/Y$ and $v=Z/Y$. The equation is $v={\prod }_j(u-{\lambda }_{j}v)$. Its derivative in $v$ at $(0,0)$ is one on the left after bringing the right side across, so $u$ is a local parameter. The equation gives $v=u^3+O(u^5)$. Consequently $x=u/v$ and $y=1/v$ have poles of orders two and three at $P_{\infty }$.

The previous permutation calculation shows that every automorphism either preserves $x^3,y^3$ separately or interchanges them. Their sum and product are therefore fixed by the Galois group and belong to $K$, proving the quadratic assertion.

For the depressed cubic, $\alpha +\beta +\gamma =0$, $\alpha \beta +\alpha \gamma +\beta \gamma =b$, and $\alpha \beta \gamma =-c$. Using $1+\zeta +{\zeta }^2=0$ givesand henceAlso $x^3{y}^3=-27b^3$. The desired polynomial isIts discriminant is $729c^2+108b^3=-27D$, where $D=-4b^3-27c^2$. Here $-27$ is a square in $K$, since $[3(\zeta -{\zeta }^2){]}^2=-27$.

One can justify the group criterion directly even when one resolvent vanishes. The nonzero root-difference producthas $d^2=D$ and changes sign exactly under odd root permutations. It is fixed by the entire group exactly when the group is contained in $A_3$. If $D$ is a square in $K$, then $d$ equals one of its two square roots in $K$; conversely, a fixed $d$ lies in $K$. ThereforeSeparability ensures $D\ne 0$, and the characteristic hypotheses ensure that all divisions and the sign argument are valid.

The irreducible cubic is separable, and its Galois group acts transitively on its three roots. The only transitive subgroups of $S_3$ areFor the cycle $\sigma :(\alpha ,\beta ,\gamma )\mapsto (\beta ,\gamma ,\alpha )$, the Lagrange resolvent for a cubic obeys $\sigma (x)={\zeta }^{2}x$. A transposition interchanging $\beta ,\gamma$ sends $x$ to $y=\alpha +\zeta \gamma +{\zeta }^2\beta$. Since $\zeta \in K$, these scalars are fixed by every automorphism. Thus the two possibilities for the image set areThese are descriptions of sets, so coincident values or a zero resolvent are not being asserted distinct.

For the ideal gas, $V=N{k}_{B}T/P$ gives ${\kappa }_T=1/P$. With $C_P/C_V=5/3$, the relation in part (ii) gives