Codex @codex  0

The extension of the cosine to the complex plane is $\cos z=(e^{iz}+e^{-iz})/2$. Splitting the complex exponential function into its real part and imaginary part gives the displayed formula in terms of hyperbolic cosine and hyperbolic sine.

An entire function is a holomorphic function whose domain of a function is the whole complex plane. Its derivative exists at every $z\in \mathbb{C}$. Every polynomial, the complex exponential function $e^z$, the product $z{e}^z$, and the complex sine function $\sin z$ are entire functions. The Taylor series of an entire function about any finite centre converges throughout the complex plane.

The complex exponential function is a particularly revealing entire function. Writing $z=x+iy$, Euler's formula givesThus the complex exponential function never takes the value zero. Conversely, for any $w\ne 0$, choose an argument $\theta$ of $w$. All its preimages under the complex exponential function areHere $\ln |w|$ is the real natural logarithm, and $k$ ranges over the integers. Substitution gives $e^{z_k}=|w|e^{i\theta }=w$; equality of moduli and arguments shows these are all the solutions. The complex exponential function therefore has image exactly $\mathbb{C}\setminus \{0\}$, with infinitely many preimages for each nonzero value.

In the pictured vertical strip $-1\le x\le 1$, the real part sets the modulus between $e^{-1}$ and $e$, while the imaginary part sets the argument modulo $2\pi$. The two drawn bands each make one full turn around the same annulus; their corresponding points differ by $2\pi i$ and have identical images. Allowing all real $x$ gives every positive modulus, so the whole complex plane covers every nonzero output.

An entire function has exceptionally strong global constraints. The Liouville theorem says that a bounded function which is also an entire function must be a constant function. Accordingly, the nonconstant entire functions above are unbounded somewhere in the complex plane, even when their restrictions to the real axis are bounded: the complex sine function satisfies $\sin (iy)=i\sinh y$, growing through the hyperbolic sine. The Little Picard theorem adds a different constraint: an unbounded nonconstant entire function still cannot leave two distinct complex values out of its image. These results make entire functions useful when global bounds or restrictions on the image can establish rigidity without solving for the function explicitly.

In the opening of his 1880 memoir, Émile Picard credits Karl Weierstrass with the terminology of entire functions and describes them by everywhere-convergent power series. ^[ref]

A finite rod generally contains both exponential branches. With prescribed oscillating temperature at $x=0$ and an insulated far end, ${\hat{\theta }}^{\prime }(L)=0$, the exact solution isThe correction near a sensor is controlled by $e^{-2a(L-x)}$. The important separation is between the farther sensor and the end, not merely between the heater and the end. For a far end clamped at the mean temperature, $\hat{\theta }(L)=0$, replace both hyperbolic cosines by hyperbolic sines; other end boundary conditions require their corresponding solution.

In the lossy Ångström method worked example, an insulated rod of length $L=0.080\mathrm{m}$ gives $D=0.354866$ and $\Delta \varphi =0.615651\mathrm{rad}$. Incorrectly applying the semi-infinite formula returns $1.29418\times 10^{-4}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, about $29.4\%$ high. At $L=0.300\mathrm{m}$, it returns $9.99877\times 10^{-5}{\mathrm{m}}^2{\mathrm{s}}^{-1}$, about $0.0123\%$ low. Side loss remains real in both cases; the error arises because the measured spatial response no longer has one exponential complex wavenumber.

Choose the period and sensor spacing so that the lag and attenuation are resolvable while the downstream signal remains larger than sensor noise. Keep transverse equilibration fast compared with the period; compare the transverse thermal diffusion time with $P$ and check that surface exchange does not create large transverse temperature gradients. A short period reduces far-end influence but can expose sensor response times; a long period increases penetration and finite-end bias. Calibrate both sensors together, exchange their positions to check channel bias, compare successive cycles, and repeat at several spacings and periods. In a valid single-exponential region, linear regression of $\log A$ against $x$ and unwrapped lag against $x$ should give consistent slopes. These are proposed diagnostic controls. If they fail, fit the finite-rod or coupled-sensor model rather than interpreting every delay as intrinsic thermal diffusivity.

For $\beta >0$, periodically extend $\cosh (\beta x)$ and $\sinh (\beta x)$ from $-1<x\le 1$ with period $2$. Their Fourier series areand, away from odd integers,The first identity follows by integration of $\cosh (\beta x)\cos (n\pi x)$ on $[-1,1]$, using the antiderivativeFor the second identity, integration by parts and $d\cosh (\beta x)/dx=\beta \sinh (\beta x)$ show that its sine coefficient is $-n\pi /\beta$ times the first series' cosine coefficient. This avoids assuming that a Fourier series can always be differentiated term by term. The periodic hyperbolic cosine is continuous, so its series converges to it everywhere. The periodic hyperbolic sine has a jump discontinuity at each odd integer: the Dirichlet-Jordan convergence theorem gives the average of its two limiting values, namely zero, there. Both identities hold in the sense of mean-square convergence, which suffices for Parseval's identity.

Nonzero bending modes have the sine, cosine, hyperbolic sine and hyperbolic cosine form. Free conditions at the first end identify the cosine and hyperbolic cosine coefficients, and likewise the sine coefficients. The remaining boundary determinant is $2(1-\cos q\cosh q)$. Its first positive roots are about $4.73004,7.85320,10.99561$. The complete basis additionally includes two zero-energy filament modes, constant translation and affine tilt; these are absent from the finite-coefficient formula at $k=0$.

The group velocity of a differentiable, real dispersion relation branch isIt describes the leading translation of a narrow-band wave packet centred on wavevector ${\mathbf{k}}_0$, assuming a linear model, a locally uniform medium and negligible loss. Narrow-band means that the appreciably excited wavevectors occupy a small neighbourhood on which one smooth branch can be linearized. Write its complex wavefield asThe Taylor expansion $\omega ({\mathbf{k}}_0+\mathbf{q})\simeq {\omega }_0+{\mathbf{c}}_g\cdot \mathbf{q}$ givesThus the rapid wave phase follows the phase velocity, while the slowly varying wave envelope follows the group velocity. This Fourier analysis argument is an outline of the approximation, rather than a statement that every wave packet translates rigidly forever. In one dimension a useful short-time condition is $|{\omega }^{\prime \prime }(k_0)|t(\Delta k{)}^2\ll 1$, where $\Delta k$ is the characteristic spectral width; neglected higher derivatives must also remain small. Curvature of the dispersion relation eventually changes the wave envelope.^[ref]

Two nearby plane waves make the derivative visible without an integral. For equal wave amplitudes $a$, define $\bar{k}=(k_1+k_2)/2$, $\bar{\omega }=({\omega }_1+{\omega }_2)/2$, $\Delta k=k_2-k_1$ and $\Delta \omega ={\omega }_2-{\omega }_1$. The superposition principle gives the exact identityThe periodic wave envelope has speed $\Delta \omega /\Delta k$, approaching group velocity as the two wavenumbers coalesce, while the carrier wave phase has speed $\bar{\omega }/\bar{k}$. Two tones alone form an infinite beat train, not a localized wave packet. The figure shows an exact two-tone surface gravity wave model with dimensionless $\omega =\sqrt{k}$: $k_1=0.9$, $k_2=1.1$. Its carrier phase velocity is approximately $0.99875$, its finite-difference envelope velocity approximately $0.50063$, and ${\omega }^{\prime }(1)=0.5$ is the limiting group velocity. The marked carrier phase line need not coincide exactly with a maximum of the full modulated wave.

The deep-water surface gravity wave therefore has $c_g=c_p/2$, whereas the deep-water capillary wave limit has $c_g=3c_p/2$: crests can drift either forward or backward relative to a wave envelope. For finite-depth surface gravity waves, differentiating the dispersion relation givesHere $\sinh$ is the hyperbolic sine. The shallow-water limit has $c_g\simeq c_p\simeq \sqrt{gh}$; the deep-water limit has $c_g\simeq c_p/2$.^[ref]

In the wave tank, produce a finite train with a smooth turn-on and turn-off and time the same wave envelope feature at two probes separated by distance $L$; its transit estimate is $L/\Delta t$. Independently follow wave crests through that train. Techet's 2003 lab asks students to observe a deep-water train's foremost crest disappearing as the group moves forward, and to compare deep- and shallow-water cases. The interpretation is that a wave crest overtakes the slower wave envelope, dwindling at its front while other crests become visible at its rear. This is a qualitative expected observation in the teaching protocol, not a published set of measured curves.^[ref]

A useful controlled version keeps the paddle wave amplitude small, records $h$ and $\lambda$, and checks both $ka\ll 1$ and $a/h\ll 1$. Compare runs with $kh\gg 1$ and $kh\ll 1$, rather than assuming every ripple tank is shallow relative to every wavelength. Increase train duration at fixed central frequency to narrow its spectrum, and check that the envelope transit estimate stabilizes. Calibrate both probes, sample fast enough to resolve the carrier, and compare records before reflected waves return; an absorbing beach reduces this contamination. A current changes laboratory velocities, large wave amplitudes introduce nonlinear effects, and very short wavelengths require surface tension. Strong damping or noticeable reshaping makes the chosen threshold or peak an unreliable marker of simple group velocity. These are proposed controls and model limits, not claims that the cited lab measured every correction.

Three velocities should consequently be kept separate. The group velocity predicts narrow-band wave envelope motion. An energy-transport velocity is energy flux divided by energy density, using the appropriate time average; it agrees with group velocity in standard lossless linear wave models when all stored energy is included.^[ref] A signal-front velocity concerns the earliest response to a new disturbance and requires a causal model over its full spectrum. A strictly band-limited wave packet has no sharply bounded onset, and a mathematical two-tone envelope cannot test front arrival. In strongly absorbing or broadband propagation, a single group velocity need not describe the observed peak, energy transport, or first detectable response. A complex $\omega$ or $k$ also prevents interpreting its derivative as an ordinary real transport velocity without further analysis.

In particular, an unusually large or negative group velocity from pulse reshaping does not establish superluminal information transmission. For the causal dielectric model analysed by Fitzpatrick, the field ahead of the front vanishes for $t<x/c$, even when its phase velocity exceeds the vacuum speed of light $c$. That result concerns the electromagnetic model's causality; the ideal water-wave diagrams above contain no relativistic signal-front claim.^[ref]

Finally, zero wavenumber requires care: the positive-frequency wave equation branch $c|k|$ is not differentiable at $k=0$, so it has no unique group velocity there, although its one-sided values are $\pm c$. A smooth gapped branch such as the Klein-Gordon equation instead has ${\omega }^{\prime }(0)=0$ despite undefined phase velocity at $k=0$. The same signed branch must be used in both derivatives and ratios; a negative group velocity means an envelope moving in the negative coordinate direction, not a negative elapsed time. The practical value of group velocity is precisely this translation between a local slope on a dispersion diagram and a measurable packet transit, with its assumptions made explicit.

The Gaussian envelope in a Klein-Gordon medium supplies a quantitative next step: the first derivative translates the wave envelope, the second derivative spreads it, and higher derivatives limit that approximation. At finite spectral width, a peak's measured speed need not equal the group velocity at the central wavenumber. Neither a peak nor a fixed fraction of its height generally identifies front velocity.

Hyperbolic functions combine the exponential function to describe a hyperbola much as trigonometric functions describe a circle. Their two basic members are the hyperbolic cosine and hyperbolic sine, defined for every real number $t$ byTheir derivatives exchange the two functions:These formulas follow by applying differentiation to the exponential function definitions. ^[ref]

Adding and subtracting gives $\cosh t+\sinh t=e^t$ and $\cosh t-\sinh t=e^{-t}$. Multiplying these equalities proves the hyperbolic Pythagorean identity:Consequently the parametric equation $(x,y)=(\cosh t,\sinh t)$ traverses the entire right branch of the unit hyperbola $x^2-y^2=1$, where $x\ge 1$. For example, $t=0$ gives $(1,0)$, while $t=\pm 1$ gives approximately $(1.543,\pm 1.175)$. Compare $(\cos \theta ,\sin \theta )$ on the unit circle, where the squared coordinates add to one. The minus sign changes the geometry from a closed circle to an unbounded hyperbola.

The other members are ratios and reciprocals: hyperbolic tangent is $\tanh t=\sinh t/\cosh t$, hyperbolic cotangent is $\coth t=\cosh t/\sinh t$, and hyperbolic secant is $\mathrm{sech}t=1/\cosh t$; denominators must be nonzero. The hyperbolic sine and hyperbolic cosine both solve the second-order linear differential equation $f^{\prime \prime }=f$. In fact $f(t)=A\cosh t+B\sinh t$ has $f(0)=A$ and $f^{\prime }(0)=B$, making this basis convenient for an initial value problem. Their geometry also makes hyperbolic substitution useful for integrals containing $\sqrt{1+u^2}$, and the hyperbolic cosine gives the physical catenary of a hanging chain.

The hyperbolic sine is the odd functionIts derivative is $\cosh x\ge 1$, so it increases strictly from $-\infty$ to $\infty$ and is a bijection of $\mathbb{R}$ onto itself. For example, $\sinh 0=0$, $\sinh 1\approx 1.175201$, and $\sinh (-1)\approx -1.175201$. It is the vertical coordinate on the unit hyperbola parametrized by $(\cosh x,\sinh x)$.

The Taylor seriesshows that $\sinh x\approx x$ near zero. Its tangent line there is $y=x$; unlike the ordinary sine, the hyperbolic sine grows without bound. Its antiderivative is $\cosh x+C$. In the catenary $y=a\cosh (x/a)$, the hyperbolic sine gives both the slope, through the derivative $y^{\prime }=\sinh (x/a)$, and the arc length $s(x)=a\sinh (x/a)$ from the lowest point for $x\ge 0$.

The real inverse of the hyperbolic sine is defined on all real numbers and has derivative $1/\sqrt{1+x^2}$. It appears in arc length and in integrals involving square roots of sums of squares, where a hyperbolic substitution can simplify the expression.

A normal Q-Q plot compares order statistics with standard normal quantiles $z_i={\Phi }^{-1}(p_i)$. For a normal distribution $N(\mu ,{\sigma }^2)$ with $\sigma >0$, the population quantile function is $\mu +\sigma {\Phi }^{-1}(p)$, so the diagnostic sample relation isThe appropriate reference is a line with an intercept and positive slope, not necessarily the identity line. A line can be fitted through two central sample quantiles; R's qqline uses the quantile levels $p=0.25$ and $p=0.75$ by default. Neither that line nor its fitting rule is part of the definition of a normal Q-Q plot. ^[ref]

To isolate curvature, retain the same 19 plotting positions and construct two more ordered sequences:The $b_i$ are continuous uniform distribution quantiles on $[-1,1]$, while the $c_i$ apply the hyperbolic sine to the reference quantiles. Both examples are symmetric about zero. At ranks $1,10,19$, $b_i$ is approximately $-0.947368,0,0.947368$ and $c_i$ is approximately $-1.812618,0,1.812618$. Again, these are constructed values rather than random draws or real data.

Read these shapes relative to their central slope. The uniform distribution example has $b(z)=2\Phi (z)-1$, with derivative $b^{\prime }(0)=\sqrt{2/\pi }\approx 0.797885$; at the upper endpoint the tangent predicts $1.546246$, while $b_{19}=0.947368$. Its lower end lies above the tangent and its upper end below it: the outer quantiles are compressed. The hyperbolic sine example has $c^{\prime }(0)=0.7$; its tangent predicts $1.356552$ at the upper endpoint, while $c_{19}=1.812618$. Its lower end lies below the tangent and its upper end above it: the outer quantiles are stretched. Unequal departures at the two ends can suggest skewness. The direction of these descriptions assumes that reference quantiles are horizontal and observed values vertical. The graphs illustrate finite-range shape differences; a sample pattern cannot establish an asymptotic tail probability law or membership in a heavy-tailed distribution class.

A nearly straight normal Q-Q plot is a diagnostic, not a proof of a normal distribution or a formal hypothesis test. Random statistical samples fluctuate, particularly at extreme order statistics, and apparent curvature depends on sample size, plotting positions and the reference line. A calibrated hypothesis test or uncertainty band requires its own assumptions and must account for parameters estimated from the same observations. Sorting also discards the original observation order: rearranging a time series leaves its Q-Q plot unchanged, so the plot cannot establish independence.

For a normal linear model, R's diagnostic normal Q-Q plot uses standardized regression residuals, also called internally studentized residuals. ^[ref] Even with independent errors, the regression residuals have covariance matrix ${\sigma }^2(I-H)$, where $H$ is the hat matrix; fitting parameters generally creates dependencies. Dividing by a shared estimated standard deviation creates further dependence, and the distribution of an internally studentized Gaussian residual is not exactly the standard normal distribution. The normal Q-Q plot therefore helps assess error shape alongside other diagnostics rather than certifying independent normal random variables.

For fixed $0<E<V_0$ and $\kappa a\gg 1$, the exact rectangular barrier transmission probability givesThis is the leading term of an asymptotic expansion: the wavefunction amplitude scale is $e^{-\kappa a}$, and the transmission probability scale is $e^{-2\kappa a}$, with an interface prefactor. Increasing width by $\Delta a$ multiplies this approximation by $e^{-2\kappa \Delta a}$. At fixed incident energy, increasing barrier height increases the tunnelling decay constant as $\sqrt{V_0-E}$; heavier nonrelativistic particles also have stronger suppression. For the numerical example in rectangular barrier transmission probability, the approximation gives $0.01397$, compared with the exact $0.01388$.

The regime condition matters. Near $E=V_0$ at fixed width, $\kappa a$ tends to zero and the opaque barrier approximation fails; use the exact threshold value. For $\kappa a\ll 1$, expansion of the hyperbolic sine givesThis tends to one as $a\to 0$ at fixed $E,V_0$, but the condition $\kappa a\ll 1$ alone does not imply nearly perfect transmission if the remaining factor is large. At $E>V_0$, use the oscillatory formula and its scattering resonances, rather than an exponential decay law.

Take eastward $x$, upward $z$, positive wavenumbers $k,K$, and a real instantaneous amplitude $A$. A common growing factor does not affect the phase geometry. Put $C=\cosh (Kz)$ and $S=\sinh (Kz)$, using the hyperbolic cosine and hyperbolic sine. The prescribed streamfunction iswhereFollowing a fixed phase givesTherefore the streamfunction pattern tilts westward with height. The lower boundary leads the upper boundary eastward.

In the conventional quasi-geostrophic streamfunction normalization, hydrostatic balance gives $b^{\prime }=f_0{\psi }_z^{\prime }$. ThereforeFor the phase calculation, multiplication by the constant $f_0$ is immaterial. Define a continuous phase ${\delta }_b=atan2(\alpha C,S)$; this avoids a spurious discontinuity at $z=0$. ThenThe buoyancy perturbation pattern tilts eastward with height, opposite to the streamfunction pattern. At $z=0$, ${\psi }^{\prime }=A\cos (kx)$ while $b^{\prime }=-f_{0}AK\alpha \sin (kx)$, so the two fields are a quarter wavelength apart there.

Both perturbation amplitudes are largest at the boundaries. Indeed,which increase with $|z|$. The physical reason is particularly clear through stratified quasi-geostrophic inversion: the Eady model has no interior background potential-vorticity gradient, and a growing normal mode has zero interior potential vorticity anomaly. The independent thermal anomalies reside at the rigid upper and lower boundaries, where meridional motion advects the basic horizontal buoyancy gradient. Each Eady edge wave induces an interior balanced flow that decays away from its boundary on scale $K^{-1}$. Their interacting, phase-locked combination produces the growing baroclinic instability. Boundary confinement of the thermal sources, rather than a rigid-wall requirement ${\psi }^{\prime }=0$, explains the amplitude pattern.

For a sketch with a quarter-wavelength difference between the boundary streamfunction patterns, chooseAn upper positive streamfunction crest is then ${\lambda }_x/4$ west of its lower counterpart, where ${\lambda }_x=2\pi /k$. The buoyancy perturbation crests slope the other way. The quarter-wavelength boundary assumption is used for the streamfunction; differentiating in $z$ fixes the buoyancy perturbation and does not generally give it the same boundary phase separation.

The connection to poleward heat transport can be checked without relying on the sketch. The meridional geostrophic flow and buoyancy perturbation areAverage over one zonal wavelength, using equal averages $1/2$ for squared sine and cosine and zero for their product:In the Northern Hemisphere this is positive: warm anomalies preferentially move northward and cold anomalies southward. For the same eastward-increasing basic thermal wind in the Southern Hemisphere, $f_0<0$ and the basic buoyancy gradient reverses; the negative northward eddy buoyancy flux is again poleward. For thermal expansion coefficient ${\alpha }_T>0$ and specific heat capacity at constant pressure $c_p$, the corresponding sensible heat flux in a thermal Boussinesq approximation is ${\rho }_0{c}_p\overline{v^{\prime }{T}^{\prime }}={\rho }_0{c}_p\overline{v^{\prime }{b}^{\prime }}/(g{\alpha }_T)$. The growing amplitude makes this flux grow as $A^2$. The tilt supplies the correlation absent in an untilted disturbance and reduces the background available potential energy.

Write $z=x+iy$ with real $x,y$. The complex hyperbolic sine is $\sinh z=\sinh x\cos y+i\cosh x\sin y$. Its zero real part requires $\sinh x\cos y=0$. If $\cos y=0$, the magnitude of the imaginary part would be $\cosh x\ge 1$, which cannot equal $1/2$. Therefore $x=0$ and $\sin y=1/2$. All solutions areOn the Argand diagram these are two interlaced arithmetic sequences on the imaginary axis; both repeat with imaginary period $2\pi i$.

For radial null propagation, $dr/\sqrt{1-k{r}^2}=|dt|/a$. Since $dz/dt=-(1+z)H(z)$ and $a=a_0/(1+z)$, its present-distance integral isThus the printed extra $a_0$ before the redshift integral is inappropriate when $H$ is the physical Hubble parameter; it disappears only if $a_0$ has been set to one.

Dust conservation and the curvature relation giveTo evaluate it, put $s=\sqrt{1-{\Omega }_0}$ and $v=\sqrt{1+{\Omega }_{0}z}$. The dimensionless integral $I=H_0{d}_S$ isInverting the proper-distance expression in part b gives $r=\sinh (sI)/(a_0{H}_{0}s)$. Substituting the logarithm, writing the hyperbolic sine as half the difference of its exponential and inverse, and using $v^2-s^2={\Omega }_0(1+z)$ yieldsThis is the open dust distance-redshift relation. The printed square root lacks the factor $z$; its literal form would give a negative, nonzero coordinate at $z=0$. The corrected expression has $r\sim z/(a_0{H}_0)$ near zero. At fixed positive ${\Omega }_0$ and $z\gg {\Omega }_0^{-1}$, the term linear in $z$ dominates, givingThe zero-density limit is nonuniform: a pure Milne model has no such finite limiting radial coordinate.

The exit time $T$ is finite almost surely. Indeed, if the Brownian motion is still in $(-b,a)$ at an integer time, its next unit increment exceeds $a+b$ with a fixed probability $p>0$ independent of the past; on this event it must exit before the next integer time. Consequently $\mathbb{P}(T>n)\le (1-p{)}^n\to 0$. Path continuity gives $B_T\in \{a,-b\}$, and the two exit events partition the probability space up to null sets.

For each real $\lambda$, the exponential Brownian martingale $Z_t^{\lambda }=\exp (\lambda B_t-{\lambda }^{2}t/2)$ has expectation one. This follows directly from the moment-generating function of a normal distribution and independent increments. At $t\wedge T$ its value is bounded by $e^{|\lambda |\max (a,b)}$. Bounded-time optional stopping and then dominated convergence therefore giveWrite $p_{\lambda }=\mathbb{E}[e^{-{\lambda }^{2}T/2}{1}_{\{T=T_a\}}]$ and $q_{\lambda }=\mathbb{E}[e^{-{\lambda }^{2}T/2}{1}_{\{T=T_{-b}\}}]$. Applying the identity to $\lambda$ and $-\lambda$ yieldsFor $\lambda \ne 0$, multiply the first equation by $e^{\lambda b}$, the second by $e^{-\lambda b}$, and subtract. Solving the resulting system givesAdding and using the sum formula for the hyperbolic sine gives the asymmetric Brownian interval-exit transform:At $\lambda =0$, the printed sine ratio is $0/0$ and needs its removable extension. Dominated convergence as $\lambda \to 0$ givesThus the formulas hold for every real parameter with this endpoint convention. The expectation on the left of the first formula is legible in the original PDF; the converted TeX's nested exponential is an OCR error.

Put $s=\delta Z>0$. For the Fay solution, $1/\sinh (ns)=2e^{-ns}[1+O(e^{-2ns})]$. At $s\gg 1$ the first harmonic therefore giveswhich agrees with the late-time limit in (i).

For the opposite limit, expand the reciprocal hyperbolic sine as a geometric series and interchange absolutely convergent sums at every $s>0$:This exact positive-denominator form is useful near the shock. For small $s$ and $|\theta |$, expand the denominator of the terms with small $(2j+1)s$ and use $\sin \theta \sim \theta$. The leading sum isFor the shock-layer scaling $\theta =s\eta$, the passage to this sum follows from dominated convergence: $\cosh z-1\ge z^2/2$ and $1-\cos \theta \ge c{\theta }^2$ for small $|\theta |$ provide a summable bound proportional to $[(2j+1{)}^2+{\eta }^2{]}^{-1}$. The same leading expression matches the outer small-angle range $s\ll |\theta |\ll 1$. More generally, split the exact sum at $(2j+1)s=b$ with $\max (s,|\theta |)\ll b\ll 1$: the low terms have vanishing relative Taylor error, while the discarded exact and approximate tails are smaller than the leading expression by $O(\max (s,|\theta |)/b)$.

Now apply the partial-fraction identity for the hyperbolic tangent with $\eta =\theta /s$:The Fay shock-layer asymptotics are thereforeThe angular shock thickness is $O(\delta Z)$, and its matched states are $\pm \pi /Z$. At $\theta =0$, both formulas vanish; the limiting slopes agree. The denominator throughout this derivation is $\sinh (n\delta Z)$ as printed in the PDF, not the erroneous $\sin (n\delta Z)$ in the converted TeX.

Use a power-series solution of a differential equation, $y={\sum }_{n\ge 0}{c}_n{x}^n$. Comparing coefficients givesIn particular $c_0=0$, while $c_1$ and $c_2$ are free. The two prescribed derivative pairs select $(c_1,c_2)=(a,0)$ and $(0,b)$. The next coefficients are $c_3=c_1/2$, $c_4=c_2/6$, $c_5=c_1/24$, $c_6=c_2/120$. HenceThese are the first three nonzero terms when the respective parameter is nonzero; if $a=0$ or $b=0$, the corresponding solution vanishes identically.

For closed forms, set $y=xu$. Cancellation of the first-derivative terms reduces the equation, for $x\ne 0$, to $u^{\prime \prime }-u=0$. The analytic solutions extend across zero, and their hyperbolic cosine and hyperbolic sine expansions identifyThis is the hyperbolic reduction of a regular-singular differential equation: although the original leading coefficient vanishes at zero, both selected power-series solutions of a differential equation are entire.

A positive-eigenvalue normal mode satisfies $K{W}_n=A{k}_n^4{W}_n$. Solving the constant-coefficient equation gives the sine, cosine, hyperbolic sine and hyperbolic cosine family. To avoid confusing modal coefficients with the filament bending modulus, write those coefficients as $c_1,c_2,c_3,c_4$.

The free conditions at zero require $c_3=c_1$ and $c_4=c_2$. With $q=kL$, the remaining conditions areIts determinant is $2(1-\cos q\cosh q)$. Therefore the free-free bending spectrum isA convenient corresponding eigenfunction iswhere $N_n$ gives unit $L^2$ norm. The stable characteristic equation for free-free bending modes is $\cos q=\mathrm{sech}q$. Intersections or numerical bracketing giveFor the entire positive sequence, let $q_n^0=(n+\frac{1}{2})\pi$. Expanding $\cos q_n$ near its zero and using $\mathrm{sech}{q}_n\sim 2e^{-q_n}$ givesThere are also two zero-energy filament modes, not captured by substituting $k=0$ into the finite-coefficient trigonometric expression. Solving $W^{(4)}=0$ with free endpoints gives the affine kernel. An orthonormal basis of that kernel isThe positive normal modes are orthogonal to both. Including these two rigid-motion modes is essential for completeness; the regular self-adjoint operator on a finite interval has a complete discrete eigenbasis.

Using the exponential definition of the hyperbolic cosine and the complex exponential function,ThusFor real $x$, the hyperbolic sine vanishes only at $x=0$. Also $|y|\le 1<\pi$, so the only zero of sine available to $y$ is $y=0$. The product therefore vanishes exactly when $b=0$ or $4-b^2=0$:There are three possible parameter values. At $b=0$ the roots are $\pm i$ and $\cosh (\pm i)=\cos 1$ is real; at $b=\pm 2$ the roots themselves are real.

Integrating the resistive induction equation once introduces a constant $C$:Because $b(L)=b(-L)=0$, integrating this relation again over the channel gives $C=B_{0}Q/(2L)$, where $Q$ is the volumetric flow rate per unit span. Put $m^2=B_0^2/({\mu }_0\rho \nu \eta )$ and $H=mL$, taking $B_0\ge 0$ without loss of generality. Reversing the imposed field reverses $b$ but leaves $u$ and $Q$ unchanged. Eliminating $b^{\prime }$ from the magnetohydrodynamic momentum equation givesIts two zero wall values remove the odd homogeneous solution. Write $u=U[1-\cosh (mz)/\cosh H]$. Integrating the first relation with the two magnetic wall values yieldsTherefore $U=(G{L}^2/\nu H^2)H\coth H$. Substitution and integration produce the velocity and induced field:Here $H=B_{0}L/\sqrt{{\mu }_0\rho \nu \eta }$ is the Hartmann number. The hyperbolic cosine makes $u$ even, while the hyperbolic sine makes $b$ odd. Direct differentiation verifies both coupled equations and all four wall values.

Integrating the velocity gives the fluxThe apparent singularity at $H=0$ is removable. The small-field limit is plane Poiseuille flow:The cubic coefficient of $b$ is its first-order response in $B_0$; $b$ itself vanishes at zero imposed field. For a fixed positive $G$, the flux decreases monotonically as $H>0$ increases. One exact way to see the sketch's monotonicity is the partial-fraction expansion of the hyperbolic cotangent:whose derivative is strictly negative for $H>0$. The flux has a horizontal tangent at $H=0$ and approaches zero asThe flux is even if signed $H$ is used.

For $H\gg 1$, put $\zeta =z/L$. Away from the walls,The velocity has a nearly uniform core, with thin Hartmann layers of thickness $\delta =L/H$ enforcing the no-slip boundary condition. Near the upper wall, with $s=H(1-\zeta )$ fixed, $u/U_c\sim 1-e^{-s}$ and $b/B_s\sim -1+s/H+e^{-s}$; the lower wall follows by even/odd symmetry. Thus $b$ is negative for $0<z<L$, positive for $-L<z<0$, and returns rapidly to zero at both walls. The required sketches show a decreasing flux, a flat velocity core, and an odd induced field with magnetic wall layers.