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Let $p_1(z)$ denote the Eulerian perturbation of a fluid variable that is the pressure amplitude, and keep $\rho$ constant. The linearized Euler equations and incompressible flow constraint areThe last equation gives $u=i{w}^{\prime }/k$. Substituting it into the first yieldsDifferentiating this formula and using the vertical momentum equation produces the Rayleigh equation for inviscid shear flow:For a growing normal mode, $c$ is nonreal, so $U-c$ never vanishes on the real $z$ axis. In each open region the base velocity is constant or linear and $U^{\prime \prime }=0$; therefore $w^{\prime \prime }-k^{2}w=0$. Decay away from the layer selects $e^{-kz}$ above and $e^{kz}$ below, while both exponentials are retained inside. Conveniently normalize these asThese are the spatially localized discrete modes relevant to instability, rather than all possible generalized neutral disturbances at a critical layer in a shear flow.

At each corner $U$ is continuous and $U^{\prime }$ jumps. The two sides have the same normal velocity and normal stress: there is no membrane, density jump or singular external force supplying a pressure jump. Thus $w$ and $p_1$ must be continuous. The pressure formula proves the required continuity of $(c-U)w^{\prime }+U^{\prime }w$. This is pressure matching at a piecewise-linear shear interface; it does not require $w^{\prime }$ to be continuous. Equivalently integrating the Rayleigh equation for inviscid shear flow through a corner gives $(U-c)[w^{\prime }]=[U^{\prime }]w$.

Write $\alpha =kd$, $y=c/U_0$, and $E=e^{2\alpha }$. Continuity of normal velocity givesAt $z=d$, the outside derivatives are $w_{+}^{\prime }=-kA$, $U_{+}^{\prime }=0$, while the inside derivatives are $w_{-}^{\prime }=-kB+kEC$, $U_{-}^{\prime }=U_0/d$. Hence pressure matching reduces toSimilarly at $z=-d$, the outside derivative is $kD$ and the inside derivative is $-kEB+kC$. Matching givesThe two equations can be written asA nonzero mode therefore requiresThis is the dispersion relation for the unbounded piecewise-linear shear layer.

With the phase convention $e^{ik(x-ct)}$, positive $\mathrm{Im}c$ gives exponential temporal growth. Since the displayed $c^2$ is real, a growing/decaying pair occurs precisely whenFor $0<\alpha \le 1/2$, $e^{-2\alpha }>1-2\alpha$ by the strict exponential tangent inequality, so every such mode is unstable. For $\alpha \ge 1/2$, define $f(\alpha )=2\alpha -1-e^{-2\alpha }$. Its derivative $2+2e^{-2\alpha }$ is positive, while $f(1/2)<0$ and $f(1)>0$. There is consequently exactly one root in that interval, andAt the cutoff the discrete roots merge at $c=0$; above it the roots are real and have no exponential growth. The physical wavelengths in the unstable band satisfy ${\lambda }_x=2\pi /k>2\pi d/{\alpha }_s\simeq 9.83d$.

This is a finite-thickness Kelvin-Helmholtz instability: long disturbances roll up the layer and draw kinetic energy from its shear. The off-diagonal $e^{-2kd}$ terms describe coupling of the two vorticity discontinuities. Short disturbances decay across the layer before strongly coupling its edges and are neutral. Indeed at $kd\to \infty$ the two phase speeds approach the velocities $\pm U_0$ of the separated edges. At $kd\to 0$, expansion of the numerator gives $c^2/U_0^2\to -1$, recovering the vortex sheet result $c=\pm i{U}_0$; the growth rate then behaves as $k|U_0|$.

For the basic axisymmetric flow, ${\mathbf{u}}_0=\Omega R{\mathbf{e}}_{\varphi }$. The steady radial Euler equation for fluid motion gives $-{\Omega }^{2}R=-p_{0,R}/\rho$; the other components give $p_{0,z}=p_{0,\varphi }=0$. Thus, in the absence of an additional body force,The additive constant is undetermined without an imposed reference pressure; the rigid boundary does not prescribe a uniform boundary pressure. Its shape is fixed by the container rather than by a free-surface pressure condition.

Let $v_R,v_{\varphi },v_z$ be the small velocity amplitudes and $W=p_1/\rho$ the pressure amplitude. Linearizing the cylindrical Euler equations for an inviscid fluid about uniform rotation, the material derivative is ${\partial }_t+\Omega {\partial }_{\varphi }$. The intrinsic frequency for the stated phase convention is thereforeThe radial and azimuthal curvature terms supply the Coriolis acceleration terms, givingFor $s\ne 0,\pm 2\Omega$, solving these equations givesThe incompressible flow constraint isOn substitution, the terms proportional to $m\Omega W_R$ cancel. The remaining equation iswhich provesThis is the pressure equation for inertial modes in a rigid rotating sphere. The exceptional frequencies used in the divisions must instead be treated in the original momentum system; they are not needed for the mode below.

A fixed rigid sphere is impermeable, so its normal velocity vanishes:Using the solved amplitudes and multiplying by $s(s^2-4{\Omega }^2)/i$ givesEquivalently divide by ${\bar{\sigma }}^2$ for $\bar{\sigma }\ne 0$. Smoothness means tangential slip is allowed: there is no no-slip condition on these inviscid normal modes. Nor is there a condition $W=0$, since the wall supplies the required varying normal pressure.

For an axisymmetric oscillation, $m=0$ and $s=\sigma$. Write $\beta =1-4{\Omega }^2/{\sigma }^2$. For $W=z(A{R}^2+B{z}^2+C)$ the radial and vertical operators areThe interior pressure equation therefore requiresThe boundary equation becomesSubstituting $R^2=r_0^2-z^2$, this polynomial must vanish for every boundary height. Its constant and $z^2$ coefficients consequently giveTogether with the interior relation the latter condition is $(4+\beta )A=0$. A nontrivial nonsingular oscillatory solution needs $A\ne 0$: if $A=0$ and $\beta \ne 0$, the other equations force $B=C=0$. Hence $\beta =-4$, givingThe cubic axisymmetric inertial mode of a rigid sphere is explicitlyIts cubic pressure is regular at the axis and centre, and its reconstructed velocity is divergence free and tangent to the sphere by the preceding identities. The frequency lies within the inertial wave band $|\sigma |<2|\Omega |$: Coriolis acceleration, rather than compression or a moving surface, supplies the restoring dynamics. The amplitude $A$ is arbitrary in this linear problem.

The cosmological perfect-fluid continuity equation gives $\rho \propto a^{-3(1+w)}$. In a flat universe the Friedmann equation then gives $H(z)=H_0(1+z{)}^{\alpha }$, where $\alpha =3(1+w)/2>0$. Writing $u=1+z$, the angular diameter distance obtained from the luminosity distance isIt is positive for $u>1$, starts at zero and tends to zero as $u\to \infty$ for every $\alpha >0$, including $\alpha <1$ where the radial comoving distance itself is unbounded. Differentiation gives the constant-equation-of-state angular diameter distance maximum:For $\alpha \ne 1$ the stationary equation is $u^{\alpha -1}=\alpha$, which has exactly one solution greater than one. For $\alpha =1$ the derivative has the sign of $1-\log u$. Hence this stationary point is always the unique maximum.

To determine the sketch's shape, differentiate $\log (1+z_m)=\log \alpha /(\alpha -1)$:with limiting value $-1/2$ at $\alpha =1$. The inequality follows because $\log \alpha -1+1/\alpha$ has its unique minimum zero at one. Thus $z_m$ decreases from infinity as $w\downarrow -1$ to $\sqrt{3}-1$ as $w\uparrow 1$. Useful points are $z_m=e-1$ at $w=-1/3$, $5/4$ at $w=0$, and $1$ at $w=1/3$. The endpoint $w=-1$ is excluded: there $d_A=z/[H_0(1+z)]$ increases to an asymptote and has no finite turning point.

First fix a genuine normalization error in the original PDF: the Gaussian exponent uses ${\sigma }^2$, so $\sigma$ must be the standard deviation, the variance is ${\sigma }^2$, and the normalized probability density function isThe printed prefactor has only $\sigma$ under its square root; its integral is $\sqrt{\sigma }$ rather than one. Use the normalized normal distribution above.

The one-scale probability of exceeding the linear spherical-collapse threshold is $P_M=\frac{1}{2}\mathrm{erfc}[{\delta }_c/(\sqrt{2}\sigma )]$. In the Press-Schechter formalism, the mass fraction above $M$ is assigned toThe factor two is necessary to recover all the mass in the hierarchical limit $\sigma \to \infty$: the uncorrected one-point upper tail tends to only one half. A precise justification is available in the excursion-set description of halo formation with a sharp-k smoothing filter. The smoothed Gaussian random field then becomes Brownian motion in its variance $S$. Reflecting a trajectory after its first crossing of ${\delta }_c$ pairs trajectories which crossed but end below the threshold with those which end above it. The Brownian reflection principle therefore gives $\mathrm{Pr}({\max }_{s\le S}\delta (s)>{\delta }_c)=2\mathrm{Pr}(\delta (S)>{\delta }_c)$.

This multiscale mass assignment is an extra modelling prescription: a Gaussian one-point probability and a collapse threshold alone do not mathematically determine a halo mass function. The “all matter” normalization additionally assumes arbitrarily large small-scale variance; a finite variance cutoff leaves an uncollapsed fraction.

Differentiate at fixed $z$ and fixed ${\delta }_c$:For decreasing $\sigma (M,z)$, a mass interval contains the positive fraction $-F^{\prime }(M)dM$. Its comoving mass is ${\rho }_{m,0}[-F^{\prime }(M)]dM$, so division by halo mass gives the Press-Schechter halo mass function:The absolute value, or equivalently a minus sign multiplying ${\partial }_M\sigma$, is essential for a positive abundance. The original PDF correctly places absolute values around the abundance and the full derivative expression; these bars are lost in the converted TeX. The density is comoving, so the prefactor uses today's matter mass density, rather than the physical density at redshift $z$.

Scale the equal intervals to unit length. A nonzero compactly supported cubic B-spline with maximum continuity is $C^2$, so its fourth derivative is a sum of impulses at the knots. Equivalently it is a combination of truncated cubes,To vanish to the right of its support, the polynomial tail must be identically zero. Expanding that tail requires ${\sum }_j{a}_j{j}^r=0$ for $r=0,1,2,3$. Four or fewer distinct knots give a full-rank Vandermonde matrix and only the zero solution. Hence a nonzero function requires at least five knots, or four intervals. At knots $0,1,2,3,4$, the one-dimensional nullspace has coefficients proportional to $(1,-4,6,-4,1)$. The unscaled combination integrates to six on $[0,4]$, so unit-integral cardinal normalization gives:This is the Cardinal cubic B-spline, nonnegative with support $[0,4]$. To obtain its cubic B-spline Bézier extraction, use local parameter $t=x-j\in [0,1]$ on interval $[j,j+1]$. The four power-form numerators, with a common divisor six, areFor $a+bt+c{t}^2+d{t}^3$, the cubic Bernstein basis coefficients are $a$, $a+b/3$, $a+2b/3+c/3$, $a+b+c+d$. Applying this conversion givesEach scalar row describes a cubic Bézier curve graph piece with abscissa controls $j,j+1/3,j+2/3,j+1$. The truncated-cube construction proves value, first- and second-derivative agreement at all joins, including the zero exterior. Minimum support and maximum continuity determine the basis up to a scalar; the displayed normalization fixes that remaining freedom.

The rule is linear in the ordinates and translation invariant on the uniform grid, so write $s(x)={\sum }_i{y}_i\beta ((x-x_i)/h)$. Find its Chaikin basis function by setting $y_0=1$ and every other ordinate to zero. The central local piece just derived has $D_0=-2$ and gives $\beta (t)=3/4-t^2$ for $|t|\le 1/2$. On the next interval, the interpolating quadratic through values $1,0,0$ at abscissae $0,1,2$ is $\frac{1}{2}(t-1)(t-2)$. Adding its limiting offset $1/8$ gives $\frac{1}{2}(3/2-t{)}^2$. Reflection gives the left interval, and locality gives zero outside. ThusThis is a centered quadratic cardinal B-spline. It is nonnegative, and its integer translates form a partition of unity because refinement preserves constant data. Its support has width three, and its first derivative agrees at the half-integer knots while its second generally jumps, consistent with the earlier calculations.

Let $N_{ij}(u,v)=B_i(u)C_j(v)$. Every available mixed derivative factors:If the univariate factors have continuity $C^r$ and $C^s$, the right-hand side has equal one-sided traces across a $u$ knot for $a\le r$, and across a $v$ knot for $b\le s$. Their products therefore remain continuous, including where knot lines cross. The basis and its surface inherit coordinate-wise continuity orders $r,s$, and joint $C^{\min (r,s)}$ continuity. In particular, degree-$p$ and degree-$q$ B-splines at simple knots give $C^{p-1}$ and $C^{q-1}$ continuity in the respective directions. Higher continuity can occur through special coefficient cancellation, but is not generally guaranteed.

Nonnegativity passes to products. Their partition of unity follows fromThus the surface remains in the convex hull of its active control points and commutes with application of affine functions. The support is the Cartesian product of the two univariate supports, giving local rectangular influence regions and efficient local evaluation. Linear precision also passes independently in each coordinate. Clamped endpoint interpolation gives boundary curves determined by the outer rows/columns and corner interpolation; parameter-line curves inherit the corresponding univariate properties. These are the relevant inherited properties, rather than a claim of arbitrary global variation-diminishing behavior for surfaces.

A parametric surface interrogation interface must specify its parameter domain, patch boundaries, trimming restrictions and any exceptional or singular points. It must evaluate $S(u,v)$ and the available derivatives, in particular $S_u,S_v$ for tangent vectors andfor the oriented normal vector where the denominator is nonzero. Second derivatives $S_{uu},S_{uv},S_{vv}$ are needed when the code requests curvature or second-order geometric error estimates; higher derivatives depend on the enquiry. The interface must report where these quantities exist, rather than assign an artificial normal at a singular parameter point.

Global searches also need subdivision into equivalent restricted patches, certified bounding volumes and a geometric approximation/error bound. These support intersection isolation, pruning and tolerance-controlled meshing. A single point-evaluation routine cannot guarantee a global search because it may miss remote branches or features between samples. Adjacent-patch topology and consistent orientation are needed to join results and interpret the represented surface.

First take $k\ge 2$, so the simple-knot B-splines are continuous, vanish at both support endpoints, and are strictly positive inside their supports. Thus the diagonal condition is $t_i<x_i<t_{i+k}$. We prove both directions of the Schoenberg–Whitney theorem, deriving the weighted zero-count fact needed for sufficiency rather than relying on an ambiguous coefficient-free hint.

For necessity, suppose $N_i(x_i)=0$. If $x_i\le t_i$, then for every row $r\le i$ and column $j\ge i$ we have $x_r\le x_i\le t_i\le t_j$, hence $N_j(x_r)=0$. The first $i$ rows have nonzero entries in at most $i-1$ columns and are linearly dependent. If instead $x_i\ge t_{i+k}$, then for $r\ge i$ and $j\le i$ we have $x_r\ge x_i\ge t_{i+k}\ge t_{j+k}$, so those entries vanish. The last $n-i+1$ rows have nonzero entries in at most $n-i$ columns and are again dependent. In either case the B-spline collocation matrix is singular. Thus invertibility implies the diagonal positivity condition.

We establish two elementary spline facts for the converse. For local linear independence of B-splines, extend the strictly increasing knots in both directions. On a cell $(t_j,t_{j+1})$ the $k$ active basis functions have indices $j-k+1,\dots ,j$. Their knot polynomials are ${\psi }_i(y)={\prod }_{\ell =1}^{k-1}(y-t_{i+\ell })$. The Cox-de Boor recurrence proves Marsden's identityHere is the induction behind that identity. For order one the indicator functions sum to one. On passing from order $k-1$ to $k$, collect the coefficient of each lower-order basis function. With $u=t_i$, $v=t_{i+k-1}$, its extra factor isThus the identity at one order multiplies by $y-x$ at the next.

The $k$ polynomials ${\psi }_i$ are independent: evaluate successively at $y=t_j,t_{j-1},\dots ,t_{j-k+1}$. The first evaluation isolates ${\psi }_j$; after its coefficient has been eliminated the next isolates ${\psi }_{j-1}$, and so on, with nonzero diagonal factors because the knots are distinct. Comparing powers of $y$ in the identity then shows that the active $N_i$ span all polynomials of degree at most $k-1$ on the cell. There are $k$ of them, so they are independent. In particular, a spline vanishes on a whole open cell exactly when all its active coefficients vanish. Padding the finite coefficient vector with zeros makes this fact apply also near the outer knots.

Next let $s={\sum }_{i=p}^q{c}_i{N}_i$ be a real spline on $I=(t_p,t_{q+k})$, with no identically zero knot cell there. The simple-knot B-spline regularity gives $s\in C^{k-2}$, and its derivatives through order $k-2$ vanish at both endpoints because it is supported on the closed interval. Suppose it has $Z$ distinct interior zeros. Its zero set initially has $Z+2$ connected components, counting the endpoints; the zeros are isolated because no cell vanishes identically.

A useful form of Rolle's theorem counts zero components, including any zero intervals that arise after differentiation. Between two successive zero components a real differentiable function is nonzero and has an interior extremum, supplying a zero of its derivative. These derivative-zero components are distinct: a zero interval connecting two such extrema would make the original function constant across an intervening zero and nonzero gap, which is impossible. If its derivative also vanishes at both endpoints, those supply two further distinct components. Thus differentiation increases the zero-component count by at least one. Apply this successively through derivative order $k-2$; every required endpoint derivative vanishes. No derivative in this process can vanish identically: that would make $s$ a global polynomial of degree below $k-1$, with all its endpoint derivatives through that degree zero, forcing $s=0$. The final derivative has at least $Z+k$ zero components.

That final derivative is continuous and piecewise affine on $q-p+k$ knot cells. Each closed affine cell can meet at most one zero component: two different zeros would force the whole cell to vanish, joining them into one component. Every zero component meets a cell, so there are at most $q-p+k$ of them. For $k=2$ this argument applies directly without differentiation. We have proved the compact-support spline zero countThis proof also accommodates zero intervals of the derivatives; counting them as infinitely many distinct zeros would invalidate a naive repeated-Rolle argument.

Now assume $N_i(x_i)>0$ for all $i$ and suppose $A_{x}c=0$ for a nonzero coefficient vector. Since $A_x$ is real, a nonzero real null vector can be chosen. Set $s={\sum }_i{c}_i{N}_i$, so $s(x_i)=0$ for all $i$. Consider one connected component of the union of the open supports of the terms with $c_i\ne 0$. If its first and last indices are $p,q$, the component is exactly $I=(t_p,t_{q+k})$. On this interval $s$ equals $s_0={\sum }_{i=p}^q{c}_i{N}_i$. Each knot cell in $I$ has at least one active nonzero coefficient, so local independence ensures $s_0$ does not vanish on any whole cell.

For every $p\le i\le q$, diagonal positivity gives $t_i<x_i<t_{i+k}$, and this places $x_i$ inside $I$. Thus $s_0$ has the $q-p+1$ distinct zeros $x_p,\dots ,x_q$, contradicting the upper bound $q-p$. There is no nonzero null vector, proving sufficiency. Together the two directions giveFor completeness, order $k=1$ consists of disjoint interval indicators, with the usual left-closed, right-open convention. An invertible ordered collocation requires exactly one site in each interval; ordering makes this equivalent to $N_i(x_i)>0$, and the matrix is then the identity. This handles possible sites at left knots without incorrectly replacing positivity by a strict interior inequality for order one.