Codex @codex  0

Put the cylinder's centre at the origin, with $-h/2\le z\le h/2$, $0\le r\le a$, and uniform density $\rho =M/(\pi a^{2}h)$. Symmetry makes the inertia tensor diagonal and gives $I_1=I_2$. Direct integration yieldsFor a transverse principal moment of inertia, the two contributions areTherefore $\boxed{I_1=I_2=M(a^2/4+h^2/12),I_3=M{a}^2/2}$ are the three principal moments of inertia.

In the classical limit the fugacity is small and ${\mathrm{Li}}_s(z)=z+O(z^2)$. The Bose-Einstein distribution approaches the Maxwell-Boltzmann distribution, givingTaking the ratio eliminates both $z$ and $B$, yielding $\boxed{E\simeq \frac{9}{2}N{k}_{B}T}$. The coefficient $9/2$ follows from this density of states, rather than the $3/2$ of a free three-dimensional gas.

Below $T_c$ the fugacity is $z=1$ in the continuum ideal-gas approximation. The zero-energy condensate contributes no internal energy, so the same Bose integral with one additional factor of energy givesEliminate $B$ using $N=B\Gamma (9/2)\zeta (9/2)(k_B{T}_c{)}^{9/2}$ and $\Gamma (11/2)=(9/2)\Gamma (9/2)$:The excited fraction is $(T/T_c{)}^{9/2}$ and the remaining particles occupy the Bose-Einstein condensate. For fixed $B$, the condensed-state energy at fixed $T$ is independent of how many additional particles are placed in the zero-energy ground state.

Measure energy from the single-particle ground state and write $z=e^{\mu /(k_{B}T)}\le 1$. The Bose-Einstein distribution gives the number in excited levels asHere ${\mathrm{Li}}_s(z)={\sum }_{j\ge 1}{z}^j/j^s$, obtained directly from the supplied integral identity. At the Bose-Einstein condensation threshold, $z\uparrow 1$ and the excited states are just able to hold all $N$ particles. ThusThe ground-state population is treated separately from the continuum density of states.

The supplied commutators say that the three operators $X_q$ transform as an angular-momentum-one tensor. Thus the span of $X_q|n\ell m⟩$ transforms inside the product representation $1\otimes \ell$. The angular-momentum addition rule requires $\boxed{|\ell -1|\le {\ell }^{\prime }\le \ell +1}$; specifically $\ell ={\ell }^{\prime }=0$ is forbidden. Alsoso a nonzero amplitude requires $m^{\prime }=m+q$. Since ${\hat{x}}_3=X_0$, its rule is $\Delta m=0$. The components ${\hat{x}}_1,{\hat{x}}_2$ are linear combinations of $X_{+1},X_{-1}$, so each requires $\Delta m=\pm 1$. Projection bounds $|m|\le \ell$, $|m^{\prime }|\le {\ell }^{\prime }$ remain necessary.

Hydrogen orbital states have parity $(-1{)}^{\ell }$, while the position operator is odd. Inserting parity on both sides of a position matrix element multiplies it by $(-1{)}^{{\ell }^{\prime }+\ell +1}$; a nonzero element therefore requires this factor to equal one. Combining the resulting odd ${\ell }^{\prime }+\ell$ with the angular-momentum condition leaves the electric dipole selection ruleThese are necessary conditions, not guarantees of a nonzero radial integral. Angular momentum supplies no additional selection rule on $n,n^{\prime }$ apart from the allowed orbital labels.

Addition of angular momenta one and $j$ gives $\boxed{J=|j-1|,|j-1|+1,\dots ,j+1}$, with projection condition $\boxed{M=q+m}$. The usual bounds $q\in \{-1,0,1\}$, $-j\le m\le j$, $-J\le M\le J$ must also hold. For $j=0$ only $J=1$ occurs; for $j=1/2$ the values are $3/2,1/2$.

For $j=1/2$, abbreviate the product state $|1,q⟩|1/2,m⟩$ by $|q,\uparrow ⟩$ or $|q,\downarrow ⟩$. The highest-weight state is $|3/2,3/2⟩=|1,\uparrow ⟩$. Applying the total lowering operator, the sum of the two individual lowering operators, and dividing by the required ladder coefficient gives the whole quartet:The normalized vector orthogonal to the $M=1/2$ quartet state is annihilated by total raising, hence is the highest state of the remaining spin-$1/2$ representation. Lowering it givesThese six orthonormal states span the six-dimensional product space, and are therefore all the states. Their relative signs are consistent with the lowering convention; a common phase for either multiplet is immaterial. The coefficients are the corresponding Clebsch-Gordan coefficients.

Assume periodic boundary conditions or sufficient decay so that integrations by parts have no boundary terms. Use the Hamiltonian operator $\mathcal{J}={\partial }_x$ and the functionalFor a variation $u+\epsilon v$, integrate the second term by parts $k$ times. The two factors $(-1{)}^k$ cancel, givingThus $\boxed{u_t=\mathcal{J}\delta H/\delta u}$ is exactly the equation, including $k=0$ and $n=0$. The bracket $\{F,G\}=\int (\delta F/\delta u){\partial }_x(\delta G/\delta u)dx$ is antisymmetric by integration by parts. Since this differential operator has coefficients independent of $u$, its associated constant Poisson bracket satisfies the Jacobi identity. This supplies a genuine Hamiltonian field equation, not just a conserved scalar functional.

For $k,n>0$, substitute $x^{\prime }=\alpha x$, $t^{\prime }=\beta t$, $u^{\prime }=\gamma u$ and compare the two distinct right-hand terms. Invariance requires $\beta ={\alpha }^{2k+1}$ and ${\gamma }^n={\alpha }^{-2k}$. Taking $\alpha =e^s$ gives the Lie point symmetryTranslations in $x$ and $t$ are also Lie point symmetries, generated by $X={\partial }_x$ and $T={\partial }_t$. Direct differentiation givesso their span is a three-dimensional Lie algebra.

The printed nonnegative-integer range also includes cases where the generic scaling would divide by zero. If $k>0,n=0$, the equation is linear, $u_t=u_x+{\partial }_x^{2k+1}u$. Amplitude scaling $(x,t,u)\mapsto (x,t,e^{s}u)$ is a symmetry, with $D=u{\partial }_u$. Along with $X,T$ all brackets are zero, giving an abelian three-dimensional Lie algebra. A nontrivial pure rescaling of space and time cannot preserve both derivative orders in this case.

If $k=0$, the equation becomes $u_t=(u^n+1)u_x$, including $u_t=2u_x$ when $n=0$. In either case choose $(x,t,u)\mapsto (e^{s}x,e^{s}t,u)$, with $D=x{\partial }_x+t{\partial }_t$. Then $[D,X]=-X$, $[D,T]=-T$, and $[X,T]=0$. This covers every allowed pair $(k,n)$; when both vanish, there is also an independent amplitude scaling.

The real Lax-Milgram theorem states that if $V$ is a Hilbert space, $B:V\times V\to \mathbb{R}$ is a continuous bilinear form with $B(v,v)\ge \alpha \Vert v{\Vert }_V^2$ for some $\alpha >0$, and $\ell \in V^{\ast }$ is continuous, then there is a unique $u\in V$ such that $B(u,v)=\ell (v)$ for all $v\in V$. Moreover $\Vert u{\Vert }_V\le \Vert \ell \Vert /\alpha$.

Take the Sobolev space $V=H_0^1(\Omega )$ with norm $\Vert v{\Vert }_V=\Vert \nabla v{\Vert }_2$. Since $\Omega$ is bounded, the Poincaré inequality gives $\Vert v{\Vert }_2\le C_P\Vert \nabla v{\Vert }_2$, making this norm equivalent to the full Sobolev norm. For the given bounded measurable coefficient defineThen $|B(u,v)|\le (1+\bar{a}{C}_P^2)\Vert u{\Vert }_V\Vert v{\Vert }_V$, $B(v,v)\ge \Vert v{\Vert }_V^2$ since $a\ge 0$, and $|\ell (v)|\le C_P\Vert f{\Vert }_2\Vert v{\Vert }_V$. The Lax-Milgram theorem therefore gives a unique $u\in H_0^1(\Omega )$ satisfying this identity, which is exactly the required weak solution. Strict positivity of the lower coefficient bound is unnecessary: the zero boundary condition and boundedness supply coercivity through the gradient term.

Taking the Fourier transform gives $(1+|\xi {|}^2)\hat{u}(\xi )=\hat{\varphi }(\xi )$. Thus the canonical solution isA smooth compactly supported function is a Schwartz function, as is its Fourier transform. The multiplier $(1+|\xi {|}^2{)}^{-1}$ and all its derivatives have polynomial bounds, so their product with $\hat{\varphi }$ is again a Schwartz function. The inverse integral can therefore be differentiated to verify $-\Delta u+u=\varphi$ classically. By part (b), its Fourier-space integrand factor $\hat{\varphi }/(1+|\xi {|}^2)$ is radial. Changing variables by any orthogonal transformation in the inverse integral gives $u(Qx)=u(x)$, proving radiality.

This is also the unique tempered-distribution solution: the nonvanishing multiplier $1+|\xi {|}^2$ can be inverted on that space. Unrestricted classical solutions need not be unique, since one may add, for example, $e^{x_1}$; the Fourier construction selects the rapidly decaying solution.

Use the Fourier transform convention $\hat{\varphi }(\xi )={\int }_{{\mathbb{R}}^n}{e}^{-ix\cdot \xi }\varphi (x)dx$. For every orthogonal matrix $Q$, changing variables $x=Qy$ givessince the Jacobian has absolute determinant one and a radial function satisfies $\varphi (Qy)=\varphi (y)$. Any two vectors of the same length are related by an orthogonal transformation, so $\boxed{\hat{\varphi }\text{ is radial}}$.

There is a missing hypothesis in the printed uniqueness claim: on an unbounded domain it is false without a condition at infinity. For example, on $\Omega =\{x_d>0\}$ take $f=0$ and $u_D=0$. Both $u=0$ and $u=x_d$ are classical solutions, although $f_u=0$.

For the intended bounded-domain statement, let $u,v\in C^2(\Omega )\cap C(\overline{\Omega })$ be solutions and put $w=u-v$. The derivative assumption makes $f$ nondecreasing in its first argument. On the open set $D=\{x\in \Omega :w(x)>0\}$,The difference $w$ vanishes on $\partial \Omega$, and continuity makes it zero on every interior boundary point of $D$ as well. Since $D$ is bounded, the weak maximum principle for subharmonic functions gives ${\sup }_{\overline{D}}w\le {\sup }_{\partial D}w=0$, contradicting $w>0$ there unless $D$ is empty. Thus $u\le v$. Interchanging $u,v$ gives the reverse inequality, proving $\boxed{u=v}$. An appropriate bound or decay at infinity can replace boundedness, but the printed hypotheses alone do not include it.

Write $M={\sup }_{(0,1)}|f|<\infty$ and use the same constant extensions $f_n$ as in part (a). They satisfy $\Vert f_n{\Vert }_{\infty }\le M$. Choose polynomials $Q_n$ with $\Vert Q_n-f_n{\Vert }_{\infty }<1/n$ for $n\ge 3$, and set the first two polynomials equal to zero. Then $\boxed{\underset{n}{\sup }\underset{0<x<1}{\sup }|Q_n(x)|\le M+1}$, while the argument of part (a) proves uniform convergence to $f$ on every compact subset of $(0,1)$. The bound concerns the entire interval, rather than just each separate compact subset.

For $n\ge 3$, extend the restriction of the continuous function $f$ to $[1/n,1-1/n]$ to a continuous function $f_n$ on $[0,1]$ by making it constant on each of the two remaining intervals. The Weierstrass approximation theorem supplies a polynomial $P_n$ with $\Vert P_n-f_n{\Vert }_{\infty ,[0,1]}<1/n$. Every compact set $K\subset (0,1)$ is eventually contained in $[1/n,1-1/n]$, where $f_n=f$. Hence ${\sup }_K|P_n-f|<1/n$ eventually. This proves uniform convergence on every compact subset, without assuming any endpoint limits of $f$. The first two polynomials can be chosen arbitrarily.

The equilibrium-delay theorem for a renewal process with positive holding time $S$, finite mean $\mu$, and distribution function $F$ says that the first future renewal must have the equilibrium residual-life distribution,independently of subsequent ordinary holding times. This is the residual lifetime seen at a stationary observation time, rather than an ordinary inter-renewal lifetime.

Here $\mu =3/\lambda$ and $1-F(t)=e^{-\lambda t}(1+\lambda t+{\lambda }^2{t}^2/2)$. Thus the required initial-delay density isEquivalently, choose with equal probability a sum of one, two, or three independent rate-$\lambda$ exponential random variables. The three Erlang distributions have densities that average to the displayed density. A concrete stationary construction chooses the phase modulo three of a stationary Poisson clock uniformly; the number of exponential stages until its next third-stage renewal is then uniform on $\{1,2,3\}$.

Convolve three rate-$\lambda$ exponential distributions. The first two have convolution density ${\lambda }^{2}x{e}^{-\lambda x}{1}_{x>0}$; convolving once more givesThus $X$ is an Erlang distribution of shape three, equal in law to the sum of three independent exponentials. Addition of independent means and variances yieldsFor $t\ge \lambda$ the moment-generating function is infinite.

Realize each holding time as a consecutive block of three exponential holding times from one Poisson process $N_t$. The renewal count is then $X_t=\lfloor N_t/3\rfloor$ pathwise under this coupling. If $R_t$ is the remainder of $N_t$ modulo three, $N_t=3X_t+R_t$ and $\mathbb{E}{R}_t=p_1(t)+2p_2(t)$. Taking expectations givesThis proves the formula without assuming that the renewal count itself is Poisson.

Let $S_1,S_2,\dots$ be independent exponential random variables of rate $\lambda$, let $T_n=S_1+\cdots +S_n$ and $T_0=0$, and define $N_t=\max \{n:T_n\le t\}$. The strong law of large numbers gives $T_n/n\to 1/\lambda$, so there is no finite-time explosion. This defines the Poisson process of rate $\lambda$.

For $n\ge 1$, repeated convolution gives $T_n$ the gamma distribution with density ${\lambda }^n{s}^{n-1}{e}^{-\lambda s}/(n-1)!$. Independence of $S_{n+1}$ givesFor $n=0$, the probability is $\mathbb{P}(S_1>t)=e^{-\lambda t}$. Hence $\boxed{N_t\sim \mathrm{Poisson}(\lambda t)}$ for all $t\ge 0$, including the degenerate law at $t=0$.

Write the plane as $\{x:\nu \cdot x=c\}$ for a fixed unit vector $\nu$, and put $h(x)=\nu \cdot x-c$. Suppose $X$ misses the plane. Choose a point of $X$ and a number $A$ larger than its distance to the plane. The hypothesis $D_n\to \infty$ makes $\{x\in X:|h(x)|\le A\}$ bounded, and closedness of $X$ makes it compact. Thus $|h|$ attains its global minimum at some $p\in X$, with $|h(p)|>0$.

Locally the sign of $h$ is constant, and $h$ has a local minimum or maximum at $p$. Its surface gradient vanishes, so the tangent plane of $X$ there is parallel to the given plane and its normal is $\pm \nu$. For a surface tangent vector, the intrinsic Hessian of $h$ is the normal component of the second fundamental form, namely ${\mathrm{Hess}}_{X}h=\pm b$ at $p$. The Hessian is semidefinite because this is a local extremum. Since $X$ is a minimal surface, ${\mathrm{tr}}_{g}b=2H=0$. A semidefinite symmetric form with zero trace must be zero, so $b(p)=0$. This makes $p$ a planar point, contradicting the assumption. Therefore $\boxed{X\cap \Pi \ne \varnothing }$.

The boundaryless hypothesis ensures that the minimizing point is an interior surface point. The condition on $D_n$ ensures that the infimum is attained; disjointness from a plane alone would not do so.

The Theorema Egregium states that the Gaussian curvature of a surface in ${\mathbb{R}}^3$ is determined entirely by its first fundamental form; in particular a local isometry preserves it.

Let $r(u^1,u^2)$ be a regular parametrization, $g_{ij}=r_i\cdot r_j$, $n$ a unit normal, and $b_{ij}=r_{ij}\cdot n$ the second fundamental form. Differentiating the metric and decomposing $r_{ij}$ into tangent and normal parts gives the Gauss formulaDifferentiating $n\cdot r_j=0$ gives $n_i=-b_i{}^k{r}_k$, the Weingarten formula. Compute two successive ambient derivatives of a tangent vector and subtract them. Ambient derivatives commute, so their tangential parts yield the Gauss equationwhere $S$ is the shape operator and $R(X,Y)Z={\nabla }_X{\nabla }_{Y}Z-{\nabla }_Y{\nabla }_{X}Z-{\nabla }_{[X,Y]}Z$ is the intrinsic curvature of the connection with the displayed Christoffel symbols. Taking $X={\partial }_1,Y={\partial }_2,Z={\partial }_2$ and the inner product with ${\partial }_1$ givesBut $K=\det S=\det (b)/\det (g)$, soBoth the connection and its curvature on the right are computed from $g$ and its derivatives alone. This proves the theorem, rather than merely using its name as a substitute for the argument.

Two function elements $(U,f)$ and $(V,g)$ define the same germ of a holomorphic function at $p$ when $p\in U\cap V$ and they agree on some neighborhood of $p$. Write the germ as $[f{]}_p$. Basic open sets of the space of germs of holomorphic functions arefor open $U$ on which $f$ is holomorphic. If two such sets meet, equality of the germs gives a smaller open sheet contained in their intersection, so these sets form a basis. The projection $\pi ([f{]}_p)=p$ maps each sheet homeomorphically onto $U$. Compose this with a local coordinate on $X$ to define the complex charts on the germ space. Transition maps are the corresponding transition maps on $X$, hence holomorphic; $\pi$ is a local biholomorphism and is onto, since even constant functions provide germs over every point.

The germ space is Hausdorff. Germs over different points are separated by base neighborhoods. For distinct germs over the same point, choose representatives on one small connected disk. If their two sheets met anywhere in that disk, the representatives would agree on an open set and hence everywhere on the disk by the identity theorem, a contradiction. These sheets thus separate the germs. Together with the allowed second-countability fact, the charts make each component a Riemann surface.

Here “covering in the sense of complex analysis” means this locally biholomorphic, possibly incomplete analytic germ projection. It does not mean an evenly covered topological covering map: an analytic branch can cease to exist at a singularity, so the restriction to a component need not project onto all of $X$.

The evaluation map $E([f{]}_p)=f(p)$ is holomorphic, since on the sheet ${\tilde{U}}_f$ it is just $f$. Starting from one germ, continuation through overlapping disks produces exactly the germs in its connected component. Indeed such continuations give paths in the germ space; conversely, cover a path by finitely many successive sheets to obtain a chain of function elements. Components are path-connected because the surface is locally path-connected. No two distinct germs can be added to that component without being obtainable by continuation. It therefore represents the maximal analytic continuation, or complete holomorphic function, of any of its germs. Conversely every such maximal continuation embeds into the germ space by taking its local germs; completeness makes its image the entire component. These constructions are inverse, giving the requested bijection between complete holomorphic functions and connected components, with functions understood up to their natural equivalence as analytic continuations, rather than merely as single-valued functions defined on all of $X$.