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A function between metric spaces is locally Lipschitz if every point has a neighbourhood on which one finite Lipschitz bound works for every pair. The neighbourhood and bound may depend on the point. A continuously differentiable real function on an open interval has this property: continuity of its derivative bounds that derivative on a sufficiently small closed interval, and the mean value theorem bounds every secant slope there.

The function $x^2$ is locally Lipschitz at every real point but has no single bound on the real line. The square root is locally Lipschitz on $(0,\infty )$, and has the bound $1/(2\sqrt{a})$ on $[a,b]$ for $0<a<b$. Adding zero to its domain changes the conclusion at that point: the ratios in the Lipschitz continuity example become unbounded, so it is not locally Lipschitz at zero. This distinction is what permits the delayed, nonunique ODE solutions illustrated under Picard-Lindelöf theorem.

The Lyapunov moment inequality says that higher positive absolute moments give no smaller absolute moment sizes on a probability space. For a real random variable $Z$ and $0<r<s$,allowing an infinite value. For $r\ge 1$ the left expression is an Lp norm; below one it is a quasi-norm, so the usual triangle inequality terminology needs that distinction. The comparison uses a probability measure of total mass one; arbitrary unnormalized measure spaces need a factor depending on their total mass.

For a nonnegative count, the first moment size is its expected value, while the second is the square root of its second moment. Both ten-trial counts in second moment method have first moment size one. The independent count has second moment size $\sqrt{1.9}\approx 1.37840$, whereas the shared-outcome count $N=10B$ has second moment size $\sqrt{10}\approx 3.16228$. The second moment distinguishes models with the same average because its square increases the influence of rare large counts.

More generally, for $N=10B$ with a Bernoulli random variable $B$ of parameter $1/10$,It rises from one at $r=1$ toward ten as $r\to \infty$. Replacing $1/10$ by a parameter $0<q\le 1$ gives $10q^{1/r}$, which decreases toward zero for a fixed positive exponent when $q\downarrow 0$, making the rare-event and large-exponent limits different questions. For a scaled Rademacher random variable taking values $-2$ and two with equal probabilities, every absolute moment size is exactly two: increasing the exponent does not distinguish signs when the absolute value is constant.

For finite $\mathbb{E}|Z{|}^s$, apply the reversed Jensen's inequality to the strictly concave function $t^{r/s}$ and the nonnegative random variable $|Z{|}^s$:Taking the positive $r$th root proves the claim. Equality in this finite case holds exactly when $|Z|$ is constant almost surely. If the higher moment is infinite, the upper bound is infinite and remains valid. Exponent zero is excluded from the statement; logarithmic or geometric limiting summaries require separately stated conventions, especially when $Z$ can vanish.

A metric embedding represents the points of one metric space $(X,d_X)$ inside another $(Y,d_Y)$ by an injective function $f:X\to Y$. The geometric question is how faithfully the representation preserves pairwise distances. For example, binary strings can be treated as vectors so that tools from linear algebra become available, but the chosen norm determines what their separation means. A drawing of the points alone does not specify a metric embedding: both metrics matter.

An isometric embedding preserves every distance exactly. A metric embedding that is a bi-Lipschitz map allows multiplicative error while keeping its bi-Lipschitz distortion bounded over all pairs. A coarse embedding instead controls separation through possibly nonlinear functions of distance. These are different requirements: the Hamming cube as an L1 and L2 metric below preserves Hamming distance exactly in one representation and takes its square root in another.

The usefulness is that a complicated distance comparison becomes a calculation with familiar vectors and norms. This also helps assess a visualization: nearby points should not acquire arbitrarily large separations, and far-apart points should not collapse together. Matoušek's 2013 lecture notes describe how motivations from computer science, especially approximation algorithms, accelerated the study of metric embeddings from the 1990s onward. ^[ref]

For a set $S$ of nonnegative integers, its natural density is the limiting proportion of the first $N$ nonnegative integers that belong to $S$:provided this limit exists. Write $C_S(N)=|S\cap \{0,\dots ,N-1\}|$ for the counting function. The numerator is a finite cardinality; the denominator is exactly $N$, including zero. Every existing natural density lies between zero and one.

For the even nonnegative integers $E=\{0,2,4,\dots \}$, there are $m$ elements below $N=2m$ and $m+1$ below $N=2m+1$. Thus $C_E(N)=\lceil N/2\rceil$ using the ceiling function, and $d(E)=1/2$. For the square numbers $Q=\{0,1,4,9,\dots \}$, the exact count for $N\ge 1$ isThe floor function counts the allowed indices $1,\dots ,\lfloor \sqrt{N-1}\rfloor$, and the extra one counts zero. At $N=10$, the square numbers below ten are $0,1,4,9$, so the proportion is $4/10$; at $N=100$, it is $10/100$. The count grows like the square root of $N$, while the denominator grows like $N$.

The prime numbers also have natural density zero. If $P$ is their set, the Prime number theorem gives $C_P(N)\sim N/\log N$, hence $C_P(N)/N\to 0$. The Prime number theorem is a substantial result about the count, rather than a consequence of the fact that there are infinitely many prime numbers. ^[ref]

Why use natural density when cardinality already measures size? The even nonnegative integers, the square numbers, and the prime numbers all have the same countably infinite cardinality, yet their proportions among large initial integer intervals differ. Natural density distinguishes frequent membership from sparse membership. Equivalently, under a uniform distribution on a finite set of $N$ possible nonnegative integers, $C_S(N)/N$ is the probability of landing in $S$. A set without natural density shows why this limiting proportion need not exist.

For $c>0$ and $a_0>0$, the displayed recurrence relation stays positive and satisfiesAfter the first step, $a_n^2\ge c$, so $a_{n+1}-a_n=(c-a_n^2)/(2a_n)\le 0$. The bounded monotone sequence theorem gives a limit of a sequence $L\ge 0$; the inequality $L^2\ge c>0$ makes $L$ positive, and passage to the limit of a sequence in the recurrence relation gives $L^2=c$. This proves existence of the positive square root without assuming it in the construction. Once $s=\sqrt{c}$ is known, the exact error identity $a_{n+1}-s=(a_n-s{)}^2/(2a_n)$ explains the eventual quadratic convergence.

The Newton root-finding iteration seeks a zero of a function $F$ by repeatedly solving its tangent line approximation. At a current guess $t_r$ with derivative $F^{\prime }(t_r)\ne 0$, setGeometrically, follow the tangent line to the horizontal axis, then return to the function graph for the next tangent line. Close to a suitable zero of a function, the error becomes dramatically smaller.^[ref]

For a concrete square root calculation, take the polynomial $F(t)=t^2-2$ and start at $t_0=1$. The Newton root-finding iteration becomes $t_{r+1}=(t_r+2/t_r)/2$, givingThe errors use the unrounded iterates. After the first step, each iterate overestimates $\sqrt{2}$, and the exact identityexhibits quadratic convergence: once close, the number of correct decimal digits roughly doubles per step. This speed makes the Newton root-finding iteration useful when accurate derivatives are available.

Global convergence requires additional care. For the polynomial $F(x)=x^3-2x+2$, starting at $x_0=0$ gives $F(0)=2$, $F^{\prime }(0)=-2$, hence $x_1=1$. Then $F(1)=1$, $F^{\prime }(1)=1$ gives $x_2=0$. The sequence repeats $0,1,0,1,\dots$ forever, although a real polynomial root lies in $(-2,-1)$ by the intermediate value theorem. Neither point in the cycle is a zero of a function, and neither has zero derivative. A maintained sign-changing interval and fallback bisection method can protect a root solver from unsuitable Newton steps.

Numerical analysis designs algorithms that approximate mathematical answers while controlling cost and error. Its central question is how much of a computed answer can be trusted. Finding a zero of a function, evaluating an integral, solving a linear system, and evolving an ordinary differential equation all demand choices about accuracy, work, and the information available. MIT's numerical-methods course explicitly treats performance and accuracy as joint concerns.^[ref]

Three questions organize error analysis. Truncation error measures the approximation introduced by an algorithm even in exact arithmetic; roundoff error comes from finite precision; the condition number measures sensitivity inherent in the problem. Numerical stability asks how the chosen algorithm propagates perturbations. These distinctions suggest different remedies: refine an approximation, change precision or arithmetic, or improve the accuracy of the input.

For example, the square root $s(a)=\sqrt{a}$ has relative condition number $|a{s}^{\prime }(a)/s(a)|=1/2$ for $a>0$: a small relative change in $a$ produces about half that relative change in $s(a)$. In contrast, $g(a)=1/(1-a)$ has relative condition number $|a/(1-a)|$. Changing $a=0.99$ to $0.9901$ changes $g(a)=100$ to about $101.010101$: an input change of about $0.0101\%$ produces an output change of about $1.0101\%$. This sensitivity exists even with exact arithmetic. MIT's conditioning lecture separates the condition number from roundoff error in precisely this way.^[ref]

Because $f(z)/z$ is holomorphic and nowhere zero on the unit disc, its square root with value one at zero exists. Defining $h(z)=z\sqrt{f(z^2)/z^2}$ gives an odd normalized univalent function. If $h(z)=h(w)$, squaring gives $z^2=w^2$; the possibility $z=-w$ and oddness force both to be zero. Applying the area theorem for univalent functions to $1/h(1/\zeta )$ yields $|a_2|\le 2$.

A pairing function is a computable function that gives a bijection from the Cartesian product ${\mathbb{N}}_0^2$ to ${\mathbb{N}}_0$, with two computable functions recovering its coordinates. Here ${\mathbb{N}}_0=\{0,1,2,\dots \}$ consists of the natural numbers, including zero. If $p$ is the pairing function and $L,R$ its decoding functions, the requirements areEvery ordered pair has exactly one code, every code represents an ordered pair, and decoding always terminates. For a bijection given by a total computable function, computability of the inverse function also follows by enumerating all ordered pairs until the one with code $z$ appears. A primitive recursive pairing function makes the stronger requirement that both encoding and coordinate decoding are primitive recursive.

The Cantor pairing function makes this concrete through triangular numbers:On the diagonal $x+y=s$, list $(s,0),(s-1,1),\dots ,(0,s)$. There are $s+1$ ordered pairs there, and $T(s)$ ordered pairs on earlier diagonals. Their codes therefore occupy precisely the consecutive natural numbers from $T(s)$ through $T(s+1)-1$. These intervals touch without overlapping and cover all natural numbers; within each interval the offset $y$ determines the ordered pair. This proves both injectivity and surjectivity, hence the bijection. It also explains why the infinite grid is a countably infinite set: every ordered pair is reached after finitely many steps. A row-by-row listing would never finish its first infinite row. Authored treatments give the same diagonal construction with the coordinates exchanged.^[ref]

For example, the Cantor pairing function givesThe order of the coordinates matters: $(1,0)$ and $(0,1)$ are different ordered pairs with different codes.

To decode a natural number $z$, find the unique diagonal index $s$ satisfyingThe inequality gives $0\le y\le s$, so both recovered coordinates are natural numbers. At $z=18$, $T(5)=15\le 18<T(6)=21$, hence $s=5$, $y=18-15=3$, and $x=5-3=2$. At $z=0$, $s=0$ and both coordinates are zero. Solving the quadratic inequality gives the equivalent floor function expressionThis is exact as a statement about the real numbers and their square roots. An implementation using floating-point arithmetic can round across a diagonal boundary, especially for large $z$. Use an exact integer square root instead: with $q=\lfloor \sqrt{8z+1}\rfloor$ computed using integer arithmetic, $s=\lfloor (q-1)/2\rfloor$. Alternatively, search for the diagonal using exact integer comparisons. The primitive recursive pairing function proof below uses bounded minimization and needs no approximate square root.

A pairing function is a reversible encoding: unlike a hash function mapping arbitrarily many inputs into a fixed finite range, it loses no coordinates. This works because the mathematical natural numbers have no upper bound. A machine word with $b$ binary digits has $2^b$ possible values, while two such words have $2^{2b}$ possible ordered pairs; a bijection into one word of the same width is impossible. Evaluating the Cantor pairing function in a fixed-width integer type can overflow, including in the intermediate product $(x+y)(x+y+1)$. Exact implementation requires sufficiently large integer storage or an explicit restricted input range. One mathematical code need not occupy fewer binary digits than its two inputs.

Iterating a pairing function gives a reversible encoding of tuples of any fixed positive length. With the left-nesting convention, the Cantor pairing function encodes $(2,3,4)$ asDecode the outer ordered pair first: $T(22)\le 257<T(23)=276$ gives $(18,4)$. Decode its first coordinate $18$ to recover $(2,3)$, giving the nested ordered pair $((2,3),4)$ and thus the original tuple $(2,3,4)$.

The tuple length and nesting convention must be known. The same code $257$ represents $(18,4)$ as a two-entry tuple and $(2,3,4)$ as a three-entry tuple; nesting alone does not make variable-length tuples self-describing. To encode variable-length tuples, supply their length or another explicit termination convention as part of the encoding. This simple change of representation is useful in computability theory and Gödel numbering: a function of several natural numbers can be studied through a single numerical input, while all coordinates remain recoverable. The striking fact is that a whole infinite grid can be serialized without omitting a point.

A historical reference for comparing sets through bijections is Georg Cantor's paper Ein Beitrag zur Mannigfaltigkeitslehre, published in 1878 in volume 84 of Journal für die reine und angewandte Mathematik, pages 242–258.^[ref]

Eliminating the shear stress from the two linear equations gives the Maxwell shear-wave dispersion equation in physical space,where $\nu$ is the kinematic viscosity. Substitution of a Fourier mode $e^{iky+\sigma t}$ givesThese are temporal exponents: the oscillation angular frequency is the imaginary part of $\sigma$, with the sign determined by the selected time convention. For $4\tau \nu k^2<1$ both exponents are real and nonpositive. At equality the repeated exponent is $-1/(2\tau )$; generic solutions include $t{e}^{-t/(2\tau )}$. Above the threshold the modes are damped propagating shear disturbances:At large $|k|$ their phase velocity tends to $\sqrt{\nu /\tau }=\sqrt{G_{\mathrm{el}}/{\rho }_0}$.

For the limit $\tau \to 0$ at fixed $\mu ,{\rho }_0$ and $k$, expansion of the square root givesAfter a short initial relaxation, the surviving branch obeys the diffusion equation $u_t=\nu u_{yy}$ of a Newtonian fluid. The fast branch disappears on the vanishing viscoelastic relaxation time; at $k=0$ the two exponents are exactly $0$ and $-1/\tau$. This limit is not uniform in $k$: modes with $k$ of order $(\tau \nu {)}^{-1/2}$ still sample the oscillatory transition. If instead $G_{\mathrm{el}}=\mu /\tau$ is held fixed as $\tau \to 0$, then $\mu \to 0$ and the fixed-$k$ slow exponent tends to zero. The ordinary Newtonian fluid limit therefore specifies which material parameter is held fixed.

Let $h=f^{-1}$, $w_0=f(z_0)$ and $z=h(w)$. Continuity of $h$ gives $z\to z_0$ as $w\to w_0$, and injectivity gives $z\ne z_0$ when $w\ne w_0$. ThereforeThe assumed nonvanishing derivative makes this limit well defined. It exists at every $w_0\in V$, so $h$ is holomorphic, with $\boxed{h^{\prime }(w)=1/f^{\prime }(h(w))}$.

Squaring maps the first quadrant bijectively onto the upper half-plane. For $\zeta$ in that half-plane, $|\zeta -i|<|\zeta +i|$, and the Cayley transform $(\zeta -i)/(\zeta +i)$ maps it bijectively onto the unit disc. Their composition isThe inverse is the unique first-quadrant square root of $i(1+w)/(1-w)$, and is holomorphic by the first argument. Hence $H(w)=g(F^{-1}(w))$ is holomorphic throughout the disc. Its Taylor series at zero converges for $|w|<1$, locally uniformly there. Substituting $w=F(z)$ gives the stated expansion, with $c_n=H^{(n)}(0)/n!$, convergent at every point of the quadrant and locally uniformly on it.

Let $h=f^{-1}$, $w_0=f(z_0)$ and $z=h(w)$. Continuity of $h$ gives $z\to z_0$ as $w\to w_0$, and injectivity gives $z\ne z_0$ when $w\ne w_0$. ThereforeThe assumed nonvanishing derivative makes this limit well defined. It exists at every $w_0\in V$, so $h$ is holomorphic, with $\boxed{h^{\prime }(w)=1/f^{\prime }(h(w))}$.

Squaring maps the first quadrant bijectively onto the upper half-plane. For $\zeta$ in that half-plane, $|\zeta -i|<|\zeta +i|$, and the Cayley transform $(\zeta -i)/(\zeta +i)$ maps it bijectively onto the unit disc. Their composition isThe inverse is the unique first-quadrant square root of $i(1+w)/(1-w)$, and is holomorphic by the first argument. Hence $H(w)=g(F^{-1}(w))$ is holomorphic throughout the disc. Its Taylor series at zero converges for $|w|<1$, locally uniformly there. Substituting $w=F(z)$ gives the stated expansion, with $c_n=H^{(n)}(0)/n!$, convergent at every point of the quadrant and locally uniformly on it.

For a nonzero eigenvector $u$ of $S^{\ast }S$,Therefore $\overline{\lambda }=\Vert Su{\Vert }^2/\Vert u{\Vert }^2$ is real and nonnegative, and so every eigenvalue $\lambda$ is real and nonnegative.

The preceding adjoint identities give $(S^{\ast }S{)}^{\ast }=S^{\ast }S$, so this linear operator is self-adjoint. The stated finite-dimensional spectral theorem supplies an orthonormal basis $e_j$ with $S^{\ast }S{e}_j={\lambda }_j{e}_j$. If $v={\sum }_j{c}_j{e}_j$, thenSince ${\lambda }_{-}\le {\lambda }_j\le {\lambda }_{+}$,Taking nonnegative square roots provesThe square roots of these extreme eigenvalues are the smallest and largest singular values of $S$; the lower bound can be zero if $S$ has a nontrivial kernel of a linear map.

Orient the contour and call its left-hand side the plus side. At a smooth interior point $t$, take a local arclength parameter $\gamma (s)$ with $\gamma (0)=t$. The Cauchy principal value means the symmetric deletion limitThe two local pieces are deleted with the same arclength cutoff. For a sufficiently smooth regular contour, an equal-radius deletion about $t$ gives the same limit. The point is assumed to be away from endpoints and self-intersections.

To derive the Sokhotski–Plemelj formula, split the contour into a short smooth arc through $t$ and its complement. The complementary integral has the same limit from both sides. On the short arc subtract $f(z)$ from the numerator. Because $f$ is holomorphic nearby, $(f(\zeta )-f(z))/(\zeta -z)$ extends continuously as $\zeta$ approaches $z$; this part also has the same limit from either side. The remaining local term is $f(z)\int d\zeta /(\zeta -z)$. Straightening the oriented tangent shows its extra contribution is $+i\pi f(t)$ from the left and $-i\pi f(t)$ from the right: on a real segment this is the direct limit of ${\int }_{-\delta }^{\delta }du/(u\mp i\epsilon )$. Dividing by $2\pi i$ yieldsIn particular the jump is $w_{+}-w_{-}=f$, while the sum is the principal-value integral divided by $\pi i$. Smooth densities on a real interval satisfy the same formulas at every interior point, by the identical subtraction step with a Hölder-continuous remainder.

Now put $h=w/\tilde{w}$. On the interval, the sign reversal of the auxiliary boundary values givesThus the required Cauchy transform density isWhenever this density is integrable at the endpoints, taking $h=C[F]$ constructs a function with the required jump. Multiplication back gives $w_{+}+w_{-}=G$. The jump of the constructed $w$ isTo conclude that this jump solves the final singular equation, one must additionally justify $w=C[f]$. This reconstruction requires endpoint and infinity conditions; the sign reversal of $\tilde{w}$ alone does not ensure it. Here is a complete valid normalization. LetThe branch is holomorphic off $[0,1]$, with $S_{+}(x)=i\sqrt{x(1-x)}$ and $S_{-}=-S_{+}$. If $G$ extends smoothly to $[0,1]$, the density $F=G/{\tilde{w}}_{+}=i\sqrt{x(1-x)}G$ vanishes like a square root at each endpoint. Its Cauchy transform $h$ is bounded at the endpoints and is $O(z^{-1})$ at infinity. Therefore $w=h/S$ is at most inverse-square-root singular at each endpoint and is $O(z^{-2})$ at infinity. Its jump $f$ is integrable.

For Cauchy reconstruction from a jump across a finite interval, apply the Cauchy integral formula to $w$ on a large slit contour, and shrink its banks to $[0,1]$. The large circle contributes zero in the limit; each small endpoint circle contributes $O(\sqrt{\epsilon })$ and also vanishes. The two banks give the jump $f$, proving $w=C[f]$. The sum part of the Sokhotski–Plemelj formula now givesThis proves the intended finite Hilbert transform inversion with admissible endpoint behavior. Adding a multiple of $1/\sqrt{x(1-x)}$ to $f$ gives a homogeneous solution; equivalently one can add a constant multiple of $1/S$ to $w$, whose boundary sum is zero.

The omitted normalization is a real limitation of the printed claim. To see it explicitly, choose instead $\tilde{w}=S$ and $G=1$. This auxiliary function satisfies all the stated holomorphicity, nonvanishing and sign-reversal requirements. ButThe displayed integral identity follows from $\xi =(1+\cos \theta )/2$ and its continuation from real $z>1$. Hence the proposed jump formula gives $f=0$, which does not solve the equation with $G=1$. The failure is the nonzero constant at infinity: $w-C[f]=1/2$. Thus the final implication is valid with suitable endpoint and decay conditions, but is false for an unrestricted auxiliary function as printed. The valid reciprocal-square-root normalization above supplies an actual solution for $G$ smooth up to the two endpoints.

Use the condensate wavefunction convention $\int |\psi {|}^2{d}^{3}r=N$, so $n(\mathbf{r})=|\psi (\mathbf{r}){|}^2$ is the number density and $\psi /\sqrt{N}$ is a normalized one-particle orbital. The contact interaction strength is $U_0=4\pi {\hslash }^{2}a/m$ for dilute atoms with s-wave scattering length $a$. The chemical potential $\mu =\partial E/\partial N$ fixes the stationary time dependence $e^{-i\mu t/\hslash }$ and enforces normalization when minimizing the energy. The Gross–Pitaevskii equation follows from an energy functional that contains the pair-counting factor $1/2$:Varying $E-\mu \int |\psi {|}^2{d}^{3}r$ gives the stationary Gross–Pitaevskii equation; the interaction contribution to that equation is $U_{0}n\psi$, without the factor $1/2$.

For the broadening discussed here assume repulsive interactions, $a>0$. The Thomas–Fermi approximation for a condensate neglects the gradient kinetic energy compared with the trapping and interaction energies. At fixed spatial scale, increasing $N{U}_0$ increases the mean-field potential. Expansion then reduces the kinetic energy further: for a cloud radius $R$, its scale per atom is ${\hslash }^2/(m{R}^2)$, whereas the trap energy scale is $m{\omega }^2{R}^2$. Their ratio decreases as $(a_0/R{)}^4$. In the weak-interaction limit the kinetic energy instead balances confinement, giving the Gaussian wave packet ground state of the quantum harmonic oscillator.

The spherical Thomas–Fermi condensate hasIntegrating this density gives $N=8\pi \mu R^3/(15U_0)$. ThusAt fixed $N$, stronger repulsion produces a wider, lower wavefunction: the Gaussian wave packet crosses over towards the square root of an inverted parabola, with an almost flat central region and a steeper outer drop. The density is the inverted parabola; the wavefunction is its square root. The exact wavefunction remains smooth and has a tail, since kinetic energy cannot be neglected near the Thomas–Fermi condensate edge rounding region. The healing length must be small compared with the cloud size in the bulk. Attractive interactions would not follow this large-$N$ expansion argument.

The original plot compares the noninteracting endpoint with two normalized repulsive spherical Thomas–Fermi condensate profiles. The latter are asymptotic shapes, rather than exact intermediate solutions of the Gross–Pitaevskii equation. Each has the same integrated atom number; $\chi =Na/a_0$ labels the illustrative interaction strengths.

Choose $A$ real and positive. The Gaussian integral givesThe three zero-point energies add for each atom. The ideal-gas chemical potential is a one-atom energy; the total energy includes all $N$ atoms.

For weak interactions, the first-order correction in perturbation theory is the energy functional of the contact interaction evaluated on the unperturbed Gaussian wave packet. The integral needed isConsequently the Gaussian interaction energy of a trapped condensate isHere $\hslash \omega ={\hslash }^2/(m{a}_0^2)$. These are the mean-field expressions consistent with the supplied equation and $\int |\psi {|}^2=N$. Exact first-order perturbation theory for precisely $N$ atoms in one orbital replaces $N^2$ by $N(N-1)$ through the number of distinct pairs; this distinction is negligible in the large-$N$ regime of the crossover estimate.

To judge the ideal-gas approximation, compare the interaction energy with the total ideal energy, or compare the interaction energy per atom with $\hslash \omega$:Thus the crossover from Gaussian to Thomas–Fermi condensate is governed by $\chi =Na/a_0$: perturbation theory is reliable for $\chi \ll 1$, while the regime of the Thomas–Fermi approximation for a condensate requires $\chi \gg 1$. The characteristic crossover number is $N_c\sim a_0/a$; there is no sharp critical atom number here. If one specifically defines crossover by $E_I=E_0$ using the weak-interaction expression, it gives $N_c=(3\sqrt{2\pi }/2)a_0/a$. That prefactor is a convention-dependent estimate extrapolated to the edge of perturbation theory, not a phase transition. Comparing $E_I$ itself to a single $\hslash \omega$ would incorrectly omit a factor of $N$.

For the numerical estimate use $m\simeq 87\times 1.66054\times 10^{-27}\mathrm{kg}$ and $\hslash \simeq 1.05457\times 10^{-34}\mathrm{J}\mathrm{s}$. ThenThe particular equality $E_I=E_0$ would instead give about $1.15\times 10^3$ atoms. These quantify different choices within a smooth crossover; a well-developed spherical Thomas–Fermi condensate profile requires going substantially beyond the characteristic scale.

In vector form the system is $U_x+A(U)U_y=C(U)$, withFor $v>0$, its two real distinct eigenvalues are $\boxed{{\lambda }_{\pm }=u\pm 2\sqrt{v}}$, with left eigenvectors ${\ell }_{\pm }^T=(1,\pm 1/\sqrt{v})$. Thus the system is strictly hyperbolic there; at $v=0$ the eigenvalues coincide and the matrix is defective, so strict hyperbolicity fails. The characteristic directions are $dy/dx={\lambda }_{\pm }$.

If $\dot{y}={\lambda }_{\pm }\dot{x}$, multiplication of $\dot{U}=U_x\dot{x}+U_y\dot{y}$ by the left eigenvector gives ${\ell }_{\pm }^T\dot{U}={\ell }_{\pm }^{T}C\dot{x}$. Since $d(u\pm 2\sqrt{v})=du\pm dv/\sqrt{v}$, defineThenConsequently $R$ is constant on the minus characteristic curves, and $S-2x$ is constant on the plus characteristic curves. One reconstructs $\boxed{u=(R+S)/2,v=(S-R{)}^2/16}$, with the essential physical constraint $S\ge R$.

For the data on $y=0$, the nonnegative square root givesThus, on any regular patch, use the two footpoints determined byand put $R=R_0(\xi )$, $S=2(x-\eta )+S_0(\eta )$, selecting branches continuous from the initial point. These implicit algebraic formulas determine the local answer wherever both characteristic curve projections are regular and $S>R$.

For example, near an initial point with $x>1$, they give $R=3/2$, $S=2x-1/2$, so $\boxed{u=x+1/2,v=(x-1{)}^2/4}$. On an initial interval $x<1$, define $H=\sqrt{9/4+4y}$. One instead has $S=H$ andnear the chosen initial point $x_0\ne 1/4$. The radicands must be positive, the footpoints must stay in their indicated initial intervals, and $H>R$ must hold.

There is no continuously differentiable solution realizing the first data set on the entire stated initial line. At $(x,y)=(1/4,0)$, $u=3/4$, $\sqrt{v}=3/8$, $u_x=1$ and $v_x=-3/8$. The minus left eigenvector is $(1,-8/3)$ and its eigenvalue is zero. The differential equation there would require ${\ell }_{-}^T{U}_x={\ell }_{-}^{T}C$, but its two sides are respectively $2$ and $0$. This is an explicit compatibility contradiction, not merely a loss of a preferred formula. Also, the proposed simple expression independent of $y$ does not solve the equation for $x<1$, because then $v_x<0$ but $\sqrt{v}>0$.

For the second data set, treat $x$ as the evolution variable and let $\eta$ label the initial $y$ coordinate. Then $R_0(\eta )=4\min (\eta ,0)$ and $S_0(\eta )=4\max (\eta ,0)$. In the positive quadrant the minus characteristic curves give $R=0$. The plus characteristic curves satisfy $S=2x+S_0(\eta )$ and $y=\eta +x^2+x{S}_0(\eta )$. Therefore their continuous characteristic curve continuation isAbove the parabola the footpoint is $\eta =(y-x^2)/(1+4x)>0$; below it the plus footpoint is $\eta =y-x^2<0$. Each branch directly satisfies both equations and $v>0$. They agree continuously on $y=x^2$, but their $y$ derivatives are $0$ and $2/(1+4x)$, respectively. Thus the continuation is piecewise classical and not continuously differentiable across the parabola. The source's globally continuously differentiable formulation needs this qualification. If only positive initial $y$ is intended, the upper rational formula is a classical continuation throughout the positive quadrant, but the region $y<x^2$ requires incoming information from $y=0$ and is not determined by that positive half of the initial line alone. This illustrates loss of differentiability from square-root hyperbolic source data.

The radial retraction of a punctured ball gives a retraction: project away from the removed interior point until reaching the unit sphere. Write $a=x_0$, $v=x-a\ne 0$, and setThe equation $|a+tv{|}^2=1$ is the quadratic equationIts constant term is negative, so it has exactly one positive root, namely $\tau (x)$. Since $|a+v|=|x|\le 1$, that positive root satisfies $\tau (x)\ge 1$. Thus $r(x)\in \partial {\mathbb{B}}^n$ for every $x\ne a$ in the closed unit ball.

The displayed expression is a continuous map on the punctured closed unit ball: its denominator is nonzero there, and its square root has strictly positive argument. If $|x|=1$, $t=1$ is the unique positive root, so $r(x)=x$. These are exactly the requirements for a retraction. The missing interior point is essential: the formula need not extend continuously to $a$, in agreement with the no-retraction theorem.

We apply the closed-bounded-domain version of the Schauder fixed-point theorem in the Banach space $L^2(\Omega )$. Write $m=|\Omega |$ and use $G,T,C_P$ from part (b)(ii). ChooseThe set $C$ is nonempty, bounded, closed and a convex set. For closedness of its positivity condition, if $w_j\ge 0$ and $w_j\to w$ in $L^2$, then pointwise $w^{-}\le |w-w_j|$, whence $\Vert w^{-}{\Vert }_2\to 0$.

The weak maximum principle already gives $Tw\ge 0$. Our estimates give, for $w\in C$,Thus $T(C)\subset C$.

To prove continuity, use the pointwise square root estimate $|\sqrt{s}-\sqrt{t}{|}^2\le |s-t|$ for $s,t\ge 0$. The Cauchy-Schwarz inequality then givesHence $T$ is a continuous map on $C$, even though the square root need not have a bounded derivative near zero.

Finally,The Rellich-Kondrachov theorem says that $H_0^1(\Omega )$ embeds compactly into $L^2(\Omega )$ on a bounded smooth domain. For $n>2$, $2<2n/(n-2)$, and for $n=2$ every finite target exponent is available, so the required compact embedding holds in every dimension in the question. Thus $T(C)$ is relatively compact in $L^2(\Omega )$.

All hypotheses of the Schauder fixed-point theorem are met. It gives $u=Tu\in C$, and part (b)(ii) converts this fixed point into the required weak solution:Only existence is needed. The construction does not assume that $\Omega$ is small or that $T$ is a contraction mapping.

Let $c=x(1_A)\ge 0$. Positivity on $a^{\ast }a$ implies positivity on every Positive element of a C-star algebra, by taking its positive square root. For a self-adjoint C-star element $h$, the positive cone of a C-star algebra givesBoth $x(\Vert h\Vert 1_A+h)$ and $x(\Vert h\Vert 1_A-h)$ are nonnegative real numbers. Their difference shows $x(h)$ is real, and the same inequalities give $|x(h)|\le c\Vert h\Vert$.

For an arbitrary $a\in A$, write $a=h+ik$ with $h=(a+a^{\ast })/2$ and $k=(a-a^{\ast })/(2i)$ self-adjoint. Since the involution is isometric, $\Vert h\Vert ,\Vert k\Vert \le \Vert a\Vert$. HenceThus every positive functional on a C-star algebra is bounded, with the preliminary bound $\Vert x\Vert \le 2x(1_A)$. This proof does not presuppose the Cauchy–Schwarz inequality for positive C-star functionals. It also gives $x(a^{\ast })=\overline{x(a)}$, since $x(h)$ and $x(k)$ are real.