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Interpret the longitudinal Dirac delta function as the Markov approximation for a random medium. The two arguments on the left of the printed covariance must be $x_1$ and $x_2$. The transverse covariance kernel is $B(|{\mathbf{z}}_1-{\mathbf{z}}_2|)$. Here statistical isotropy means transverse isotropy: a medium with a distinguished longitudinal white noise direction and a smooth transverse covariance is not literally isotropic in all three directions. Also, ideal Gaussian white noise replaces the original finite-variance field; it cannot simultaneously satisfy a pointwise normalization $\mathbb{E}{W}^2=1$.

Write $Q(r)={\mu }^{2}B(r)$ for the covariance strength of the refractive index fluctuations. Define the transverse power spectrum using the unnormalized forward Fourier transform printed in equation 3:This is the spectrum of the fluctuations, excluding the deterministic mean index. Equivalently it is the zero-longitudinal-frequency slice of the three-dimensional fluctuation spectrum before the Markov approximation. The angular integral is $2\pi J_0(\nu r)$, where $J_0$ is the Bessel function of the first kind. Consequently Fourier inversion and the Hankel transform giveThe printed equations 4 and 5 omit these reciprocal factors, and the left-hand side of equation 5 should depend on $r$. They are instead a consistent order-zero Hankel transform pair if their $F$ denotes $S_H=S_F/(2\pi )$.

Using $J_0(0)=1$, the mean-field solution in the Fourier transform convention isHere $U_x$ is the Fresnel propagator. Finiteness of the integral ensures a finite screen wave phase variance. If the quoted power spectrum uses the self-reciprocal Hankel transform convention, the same result readsThese formulas describe identical media when $S_F=2\pi S_H$; assigning the same numerical function to both spectral conventions describes different covariance strengths.

The transition equation reduces to the Bessel differential equation because $d/dT=-Zd/dZ$ and $d^2/d{T}^2=Z^2{d}^2/d{Z}^2+Zd/dZ$. Thus $Y_{TT}+e^{-2T}Y=0$ becomes $Z^2{Y}_{ZZ}+Z{Y}_Z+Z^{2}Y=0$.

The Bessel function of the first kind has $J_0(Z)=1+O(Z^2)$ at zero, while the Bessel function of the second kind has $Y_0(Z)=(2/\pi )(\log (Z/2)+\gamma )+O(Z^2|\log Z|)$. Since $\log Z=-T$, these yield a constant and a linear function of $T$, explaining the late-time behavior. At large positive $Z$,Matching $A_0\cos (Z-\pi /4)+B_0\sin (Z-\pi /4)$ to $\sqrt{\pi /2}\sin ({\epsilon }^{-1}-Z)$ fixes $A_0=\sqrt{\pi /2}\sin \theta$ and $B_0=-\sqrt{\pi /2}\cos \theta$. The large- and small-argument asymptotic expansions thereby connect the oscillatory WKB approximation to the nonoscillatory late-time solution through one Bessel transition for an exponentially decaying oscillator.

First remove the small damping term by writingThe transformed linear ordinary differential equation isThe leading frequency is $\omega =e^{-\epsilon t}$. Its WKB approximation has amplitude ${\omega }^{-1/2}$ and phase ${\int }_0^t\omega ds=(1-e^{-\epsilon t})/\epsilon$. The initial conditions selectThis leading expression has $y(0)=0$ and $y_t(0)=1$. Its WKB approximation for a slowly varying oscillator requires $|{\omega }_t|/{\omega }^2=\epsilon e^{\epsilon t}\ll 1$, so it fails around $t_{\ast }={\epsilon }^{-1}\log ({\epsilon }^{-1})$.

To resolve the Bessel transition for an exponentially decaying oscillator, shift to $T=\epsilon t-\log ({\epsilon }^{-1})$, put $Z=e^{-T}$ and rescale $y={\epsilon }^{-1/2}{e}^{-1/4}Y(T)$. The exact transformed equation isFor fixed $T$ its leading form is $Y_{TT}+e^{-2T}Y=0$. The substitution $Z=e^{-T}$ turns it into the order-zero Bessel differential equation, so $Y=A_0{J}_0(Z)+B_0{Y}_0(Z)$. In the overlap $1\ll Z\ll {\epsilon }^{-1}$, matching the large-argument Bessel functions to $Z^{-1/2}\sin ({\epsilon }^{-1}-Z)$ gives, with $\theta ={\epsilon }^{-1}-\pi /4$,Here $J_0$ and $Y_0$ are the Bessel function of the first kind and Bessel function of the second kind; the symbol $Y_0(Z)$ in this question is unrelated to the leading inner function in the preceding question.

At late times $T\gg 1$, the small-argument expansions givewhere $\gamma$ is the Euler--Mascheroni constant. The solution becomes asymptotically linear rather than maintaining the exponentially growing WKB envelope. Its leading late-time slope is ${\epsilon }^{1/2}{e}^{-1/4}\sqrt{2/\pi }\cos \theta$. These are leading asymptotic coefficients as $\epsilon \to 0$: near a zero of $\cos \theta$, higher-order phase corrections determine the small actual slope. The formula is not an absolute-error estimate uniform to arbitrarily late times. The exact late slope of an exponentially damped oscillator, obtained from a Kummer function and a Wronskian, provides a separate check even near those exceptional phases.

For $y^{\prime \prime }+P(x)y^{\prime }+Q(x)y=0$, a singular point $x_0$ is a regular singular point if $(x-x_0)P(x)$ and $(x-x_0{)}^{2}Q(x)$ are analytic there. The Bessel differential equation has $xP=1$ and $x^{2}Q=x^2-{\nu }^2$, so zero is regular singular.

Use the Frobenius method with $y=x^s{\sum }_{n\ge 0}{a}_n{x}^n$, $a_0\ne 0$. Matching coefficients gives $(s^2-{\nu }^2)a_0=0$ and $[(s+n{)}^2-{\nu }^2]a_n+a_{n-2}=0$, with missing negative-index coefficients zero. The regular root is $s=\nu$. Then $a_1=0$, all odd coefficients vanish, and the Frobenius series for Bessel functions has recurrenceFor noninteger $\nu$, the root $s=-\nu$ supplies an independent even-power series $y_2=x^{-\nu }{\sum }_{m\ge 0}{b}_{2m}{x}^{2m}$ with $b_{2m}=-b_{2m-2}/[4m(m-\nu )]$ and $b_0\ne 0$. None of these denominators vanishes. With standard normalization this is $J_{-\nu }(x)$, independent of the Bessel function of the first kind $J_{\nu }(x)$. The real powers here are understood for $x>0$.

For integer $\nu \ge 0$ and the stated normalization $a_0=1/(2^{\nu }\nu !)$, iteration givesThe ratio of consecutive term magnitudes is $|x{|}^2/[4(m+1)(m+\nu +1)]\to 0$. Thus the series converges absolutely for every finite complex $x$, defining an entire function for integer $\nu$. At integer order, $J_{-\nu }=(-1{)}^{\nu }{J}_{\nu }$; an independent solution is instead the Bessel function of the second kind.

Introduce a Lagrange multiplier for the normalization and vary $J[y]={\int }_0^1[p(y^{\prime }{)}^2+q{y}^2-\lambda w{y}^2]dx$. With variation $\eta$ satisfying $\eta (1)=0$, integration by parts givesThe upper endpoint term vanishes, and the lower term vanishes because $p(0)=0$ with the stipulated regular differentiable behavior. More generally the natural endpoint condition is $p{y}^{\prime }\to 0$ there. The fundamental lemma of the calculus of variations gives $(p{y}^{\prime }{)}^{\prime }-qy+\lambda wy=0$, the stated Sturm-Liouville problem.

The Rayleigh-Ritz variational principle estimates the lowest eigenvalue by minimizing $R[y]=\int [p(y^{\prime }{)}^2+q{y}^2]dx/\int w{y}^{2}dx$ over an admissible finite-dimensional trial family. Restricting the family can only raise the minimum.

For the specific radial problem, $p=w=x$ and $q=0$. The trial $y=A(1-x^2)$ has ${\int }_0^{1}x{y}^{2}dx=A^2/6$. Normalization gives $\boxed{A=\sqrt{6}}$, with the negative sign an equivalent choice. The numerator is ${\int }_0^{1}x(2Ax{)}^{2}dx=A^2$, soThis is the quadratic case of the power-law variational bound for a circular membrane. The exact first eigenvalue $5.7815$ is smaller because the polynomial trial is not the exact Bessel function of the first kind eigenfunction. Expanding a trial in the exact orthonormal eigenbasis expresses its Rayleigh quotient as a weighted average of eigenvalues, so no admissible trial can lie below the lowest one.

Write the analytic solution as $w(z)={\sum }_{n\ge 0}{a}_n{z}^n$. The constant term in the differential equation gives $a_1=0$, and matching the coefficient of $z^{n+1}$ givesAll odd coefficients therefore vanish; $a_2=-a_0$ and $a_4=a_0/4$. The first three nonzero terms areIn full, $a_{2j}=(-1{)}^j{a}_0/(j!{)}^2$, so $w=a_0{J}_0(2z)$, a Bessel function of the first kind. The ratio test proves convergence for every complex $z$. The other independent solution has a logarithmic singularity at zero and does not satisfy the analyticity requirement.

Let $Y=\hat{y}$. Transform the three terms of the order-zero Bessel differential equation using part (a):Adding and dividing by $i$ givesOn $-1<\omega <1$, separation of variables gives $Y^{\prime }/Y=\omega /(1-{\omega }^2)$, hence $Y=C/\sqrt{1-{\omega }^2}$. To obtain the usual solution regular at $t=0$, choose the compactly supported extensionThis is a genuine distributional weak solution: the differential equation can be written $[(1-{\omega }^2)Y{]}^{\prime }+\omega Y=0$, and $(1-{\omega }^2)Y=C\sqrt{1-{\omega }^2}{1}_{(-1,1)}$ vanishes continuously at both endpoints. Thus differentiation produces no endpoint Dirac delta terms.

Fourier inversion now givesThis integral is smooth at the origin and solves the original equation. Its value there is $C/2$. Normalizing $y(0)=1$ fixes $C=2$, and identifies the Bessel function of the first kind of order zero:The last equality uses $u=\cos \theta$. Without normalization, the regular solution is an arbitrary constant times this integral; this is the Fourier transform of the Bessel function J0 pair.

Regularity qualification. The source specifies the differential equation but no condition at its singular point $t=0$. The transformed equation separately has solutions $C_{\pm }/\sqrt{{\omega }^2-1}$ on the two exterior intervals. Thus it does not, by itself, determine the support or a single global multiplicative constant. The compactly supported choice above constructs the regular Bessel branch. The other independent branch, the Bessel function of the second kind, is singular at the origin. Also $J_0$ is not absolutely integrable on the real line, so its transform is properly taken as a tempered distribution, rather than relying on the stronger classical integrability assumptions of part (a).

The fundamental nonnegative mode is ${\varphi }_0{J}_0(2.405r/R)\cos (\pi z/H)$. Put the origin at the centre of the cylinder, so $-H/2\le z\le H/2$. Neglecting extrapolation imposes zero neutron flux at $r=R$ and $z=\pm H/2$. WriteFor an axisymmetric field, separation of variables in cylindrical coordinates givesChoose the axial separation constant so $Z^{\prime \prime }+{\beta }^{2}Z=0$. Midplane symmetry and the lowest positive axial mode give $Z(z)=\cos (\beta z)$, withThe radial equation isThe solution regular on the axis is the zero-order Bessel function of the first kind, $F(r)\propto J_0(\alpha r)$; the independent second-kind solution is singular there. The first radial zero gives $\alpha R=j_{0,1}\simeq 2.405$. Higher zeros would introduce sign changes and do not describe the fundamental nonnegative neutron flux. Normalize by the central value, using $J_0(0)=1$:The printed $2.405$ is a rounded Bessel function zero, rather than an exact root.

Each mutually incoherent patch of an extended star produces a displaced double-slit interference pattern. The detector adds their light intensities. For two apertures separated by a projected interferometric baseline $D$, a source direction offset by a small angle ${\vartheta }_x$ changes their phase difference by $2\pi D{\vartheta }_x/\lambda$. The normalized sum over the source's angular brightness distribution is its first-order optical coherence, whose magnitude is the interferometric visibility for equally bright channels.

For a uniform circular stellar disk of full angular diameter $\alpha$, the normalized angular Fourier transform gives the uniform-disk stellar visibilityHere $J_1$ is the order-one Bessel function of the first kind. This follows by integrating $e^{i(2\pi D/\lambda ){\vartheta }_x}$ over the disk and dividing by its area. In particular, its power series gives $2J_1(x)/x=1-x^2/8+O(x^4)$.

For an ordinary slit separation $d=1\mathrm{mm}$, using the paper's $\alpha =22.6\times 10^{-8}\mathrm{rad}$ and observing wavelength $\lambda =570\mathrm{nm}$,Even $d=10\mathrm{mm}$ gives a visibility loss of only $1.94\times 10^{-5}$. Thus millimetre separation leaves the interference fringes almost indistinguishable from those of a point source; it gives very little sensitivity to the diameter.

The movable mirrors in the Michelson stellar interferometer collect light at a large adjustable separation $D$ and redirect it into closely spaced output beams. They increase the sampling interferometric baseline independently of the fixed output slit spacing $d$: $D$ sets sensitivity to source structure, whereas the output optics set the screen's interference fringe spacing. The first Bessel function zero is $x_1=3.83170597$, henceThe interference fringes disappear at this baseline even though each beam separately remains bright.

The value $3.08\mathrm{m}$ assumes a uniform disk. The original question supplies a diameter without a surface brightness law; real limb darkening changes the precise visibility zero. A one-dimensional uniformly bright strip of the same full angular width instead has a sinc function visibility and first zero $D=\lambda /\alpha =2.522\mathrm{m}$. That is a different source model, not the circular-disk result. The robust conclusion without a surface brightness model is that a baseline of a few metres, rather than millimetres, is required.

Set $t=\sqrt{\lambda }x$ and $w(t)=y(t/\sqrt{\lambda })$. The chain rule turns the equation into the parameter-one Bessel differential equation. By the regular-at-zero solution property supplied in the paper,For a nonzero solution, $c\ne 0$, so $y(1)=0$ forces $J_k(\sqrt{\lambda })=0$. Conversely, every positive zero $z$ supplies the admissible eigenfunction $J_k(zx)$ with eigenvalue $z^2$.

Here is a direct justification of the infimum characterization that does not assume a minimizer exists. Let $u(x)=J_k(z_{k,1}x)$ and choose its sign so that $u>0$ on $(0,1)$. The first positive zero exists by the oscillatory large-argument behavior of the Bessel function of the first kind. At zero, its Frobenius series for Bessel functions gives $u(x)=C{x}^k(1+O(x^2))$. At one, its zero is simple by uniqueness for a regular second-order ordinary differential equation. Expanding $y=u(y/u)$ and using the equation for $u$ gives the Ground-state representation of a Sturm-Liouville energy:For completeness, apply the identity first on $[\epsilon ,1-\epsilon ]$. The discarded boundary term is $[x{u}^{\prime }{y}^2/u]$. It tends to zero at both ends: $x{u}^{\prime }/u\to k$ and $y=O(x)$ at zero, while $u^{\prime }/u=O((1-x{)}^{-1})$ and $y=O(1-x)$ at one. Taking $y=u$ achieves equality, soThe infimum is over the stated smooth admissible functions with finite energy; the equality also proves attainment by a multiple of $u$.

The incident plane wave is periodic in $\theta$, so expand it as a Fourier series ${\sum }_m{R}_m(r)e^{im\theta }$. Substitution into the Helmholtz equation, with $\rho =kr$, gives the radial Bessel differential equationRegularity at the origin selects the Bessel function of the first kind, rather than the singular second solution. ThusThe coefficients $a_m$ are independent of $r$; their values follow from matching near the origin in the next part. The completed identity is the Jacobi-Anger expansion.

In the unit half-cylinder $z\ge 0$, specifying only the curved-side boundary values does not determine a bounded harmonic function. The displayed nonzero function, where $j_{0,1}$ is a positive zero of the Bessel function of the first kind $J_0$, is regular at the axis, bounded, harmonic, and zero on $r=1$. It may be added to any particular solution without changing the side data, and even decays at infinity. Boundary information at the base $z=0$ is still required for uniqueness.

A single-tone FM signal has components displaced from the carrier wave frequency by integer multiples of the message frequency. These are temporal sidebands; the spatial sidebands used to perturb a spatial pattern are a related but different setting. To find the temporal components, apply the Jacobi-Anger expansion to the complex representation:Here $J_k$ is a Bessel function of the first kind. The identity follows by setting the generating variable to $e^{\mathrm{i}\theta }$ in its generating function.^[ref]

Taking the real part gives the exact real signal expansionThe complex representation has a Fourier coefficient $A{J}_k(\beta )$ at $f_c+k{f}_m$; the normalized magnitude is $|J_k(\beta )|$. The identity $J_{-k}(\beta )=(-1{)}^k{J}_k(\beta )$ makes upper and lower complex sideband magnitudes equal, but their signs matter when reconstructing the wave phase. A negative coefficient reverses a component's wave phase by $\pi$; an amplitude display alone hides that information.

For $\beta =5$, original evaluation gives $J_0(5)\approx -0.17760$, $J_1(5)\approx -0.32758$, $J_5(5)\approx 0.26114$ and $J_6(5)\approx 0.13105$. Thus there are appreciable components at $14$ and $26\mathrm{Hz}$, outside the $15$–$25\mathrm{Hz}$ range of the instantaneous frequency. There is no contradiction: instantaneous frequency describes the local rate of the combined wave phase, while the frequency spectrum describes a global superposition of constant-frequency components.

For the real signal, negative frequencies must also be accounted for: a cosine at $-f$ equals one at $f$. When $f_c/f_m=20$, the normalized positive-frequency cosine coefficient at $n{f}_m$, $n>0$, is exactly $J_{n-20}(\beta )+J_{-n-20}(\beta )$. The second term is extremely small over the plotted range, but the figure includes it. The chosen complex representation need not have a frequency spectrum supported only at positive frequencies, even when its specified instantaneous frequency is everywhere positive.

For a nonzero FM modulation index, the ideal frequency spectrum has infinitely many sidebands. Many become tiny, but drawing only visible lines does not make the exact signal bandwidth finite. A useful engineering width must state a threshold or a retained-power fraction. For the complex representation, Parseval's identity gives a simple check:At $\beta =5$, the sum through $|k|\le 6$ is approximately $0.99356$. Retaining those complex components captures over $99\%$ of their squared magnitudes and spans $12\mathrm{Hz}$ from $14$ to $26\mathrm{Hz}$. This is a stated truncation criterion, not an exact cutoff or a universal bandwidth rule. It also explains why a diminished carrier component need not imply a diminished envelope: its contribution has been redistributed among sidebands.

The spherical Bessel functions solve the radial wave equation in three dimensions. In particular,The apparent singularity at the origin is removable, as follows from the power series of the Bessel function of the first kind.

The dimensionless waveguide parameter $V$ combines radius, vacuum wavelength and the numerical aperture factor. In the weak-guidance scalar model the first additional spatial family appears at $V=j_{0,1}=2.404825\dots$, the first positive zero of a Bessel function of the first kind. The fundamental family has no positive cutoff. Its two polarizations are not two distinct spatial profiles.

Stipulate a distant circular source of uniform angular brightness, full angular diameter $\theta \ll 1$ in radians, and narrow optical bandwidth around wavelength $\lambda$. Distinct source patches emit independently. Assume each collector is small enough not to resolve the disk individually. For a sky offset $\mathbf{\alpha }$ from the disk centre, the baseline-dependent phase difference is $2\pi {\mathbf{b}}_{\perp }\cdot \mathbf{\alpha }/\lambda$, up to channel-sign convention. Summing independent patch contributions makes the equal-delay first-order optical coherence the normalized Fourier transform of angular brightness:Choose the horizontal coordinate along ${\mathbf{b}}_{\perp }$, put $k=2\pi B/\lambda$, and integrate along vertical chords of the disk:The imaginary part cancels between opposite chords. Here $J_1$ is the order-one Bessel function of the first kind. One way to obtain the last equality is to integrate the exponential's power series term by term on the finite disk. Equivalently the result is specified by the convergent seriesAt $B=0$ take the continuous limit $\gamma =1$; $V^2=|\gamma {|}^2$ starts as $1-x^2/4+O(x^4)$. This is why very short interferometric baselines give little leverage on angular diameter. Larger disks lose spatial optical coherence at shorter baselines.

The first positive zero of $J_1$ is $x_1\approx 3.831705970$, soThis is a full diameter convention. The angle must be converted before substitution: one arcsecond is $\pi /648000$ radians, and one milliarcsecond, abbreviated mas, is $10^{-3}$ arcseconds. For the stipulated $\theta =5\mathrm{mas}$ and $\lambda =500\mathrm{nm}$, $\theta =2.4240684\times 10^{-8}\mathrm{rad}$ and $B_0\approx 25.1575\mathrm{m}$. At $B=10\mathrm{m}$, the model gives $x\approx 1.52309$ and $V^2\approx 0.542783$.

Beyond $B_0$, $\gamma$ changes sign but $V^2$ stays nonnegative and develops weak further lobes. A single measured squared visibility can therefore match several angular diameters; sampling several interferometric baselines constrains the fit. Limb darkening, spectral averaging, or resolved collector apertures change the uniform-disk prediction. A uniform-disk diameter is a parameter of that brightness model; converting it to a physical stellar radius also requires distance and an appropriate brightness profile.