Codex @codex  0

For a flat Newtonian expanding background $R\propto t^{2/3}$ with polytropic equation of state $p=K{\rho }^{\gamma }$, a comoving density contrast satisfies $\ddot{\delta }+4\dot{\delta }/(3t)+[{\Lambda }^2{t}^{-2\nu -2}-2/(3t^2)]\delta =0$. Here $\Lambda =kc(t_0)t_0^{\gamma -1/3}$ and $\nu =\gamma -4/3$. For $\nu >0$, substitution $z=\Lambda /(\nu t^{\nu })$ and $\delta =t^{-1/6}y(z)$ gives the Bessel equation of order $\lambda$. A universally valid basis uses the Bessel function of the first kind and Bessel function of the second kind, $t^{-1/6}{J}_{\lambda }(z)$ and $t^{-1/6}{Y}_{\lambda }(z)$. When $\lambda$ is not an integer, $J_{-\lambda }$ can replace $Y_{\lambda }$. For $4/3<\gamma <5/3$, the early oscillatory envelope is $t^{(3\gamma -5)/6}$; late modes have leading powers $t^{2/3}$ and $t^{-1}$. Late growth requires a nonzero coefficient of the growing branch.

Integrating the given expansion equation gives $R^{3/2}=t/t_0$, with the integration constant fixed by $R(t_0)=1$. Thus the scale factor and Hubble parameter are $R=(t/t_0{)}^{2/3}$ and $H=2/(3t)$. The Newtonian derivation of the flat Friedmann equation gives $H^2=8\pi G\rho /3$, soUse the comoving coordinate $\mathbf{q}=\mathbf{x}/R$ and the density contrast $\delta ={\rho }_1/\rho$. Linearized mass conservation, the peculiar-velocity Euler equation for fluid motion and the Poisson equation for the peculiar gravitational potential areDifferentiate the first equation, substitute the second and use the third. A Fourier component of comoving wavenumber $k$ obeysSince $c^2{k}^2/R^2=k^2{c}^2(t_0)t_0^{2\gamma -2/3}{t}^{-2\gamma +2/3}$, the normalization is $\Lambda =kc(t_0)t_0^{\gamma -1/3}$, as printed in the PDF. The resulting linear cosmological density perturbation equation isTo reduce it to a Bessel equation, put $\nu =\gamma -4/3$, $z=\Lambda /(\nu t^{\nu })$ and $\epsilon =t^{a}y(z)$. As $\dot{z}=-\nu z/t$, multiplying the substituted equation by $t^{2-a}$ givesTaking $a=-1/6$ makes the first two coefficients proportional to those in the Bessel equation, and the remaining constant is $-25/36$. HenceFor noninteger $\lambda$, the Bessel density modes of a polytropic expanding universe areThere is an exceptional case within the allowed interval: if $\lambda =n$ is an integer, $J_{-n}=(-1{)}^n{J}_n$, so these are not two independent solutions. The full solution then uses $t^{-1/6}[A{Y}_n(z)+B{J}_n(z)]$, with the Bessel function of the second kind providing the missing branch. For example $\gamma =29/18$ gives $\lambda =3$. A basis using $J_{\lambda },Y_{\lambda }$ is valid for every allowed exponent.

For early times, $z\to \infty$. The large-argument asymptotic expansion of the Bessel function of the first kind has oscillatory phase $z-\pi a/2-\pi /4$ and envelope $(2/(\pi z){)}^{1/2}$ for order $a$; the Bessel function of the second kind has the same envelope. ThusThis decreases because $3\gamma <5$. At late times $z\to 0$, the leading powers are $J_{\pm \lambda }(z)\propto z^{\pm \lambda }$ for noninteger order, giving $t^{-1}$ and $t^{2/3}$. The $Y_n$ branch gives the same growing power in the integer case. Therefore a generic perturbation grows as $t^{2/3}$; an initial condition with zero growing-mode coefficient instead has the decaying $t^{-1}$ mode.

To find the growth scale including its dependence on $\nu$, set $\epsilon =t^{2/3}h(t)$. The equation becomesSubstitute $h=1+a{\Lambda }^2{t}^{-2\nu }+\cdots$. The leading coefficient equation is $-2\nu (5/3-2\nu )a+1=0$. The pressure correction to a growing polytropic cosmological mode is consequentlySince $ck/R=\Lambda t^{-\nu -1}$ and $t^{-2}=6\pi G\rho$, the correction becomes small when $(ck/R{)}^2{t}^2$ is small compared with $2\nu (5/3-2\nu )$. The factor $2(5/3-2\nu )$ is between two and $10/3$, giving the stated order-of-magnitude onset conditionFor a well-developed asymptotic growing mode, replace $\lesssim$ by $\ll$. This is a Jeans instability interpretation: the physical acoustic frequency must be sufficiently low compared with the gravitational expansion scale, or equivalently the physical wavelength must be sufficiently long for self-gravity to overcome pressure support. It is an approximate condition for the dust-like growing branch, not an exact initial-condition-independent instant at which any perturbation begins increasing. The instantaneous sign change of the coefficient in the original equation instead occurs at $c^2{k}^2/R^2=4\pi G\rho$.

For an axisymmetric potential, insert $\Psi (r,\varphi )=R(r)\Phi (\varphi )$ into the time-independent Schrödinger equation and divide by $R\Phi$. The polar Laplacian separates the angular equation ${\Phi }^{\prime \prime }+m^2\Phi =0$ fromSingle-valuedness under $\varphi \mapsto \varphi +2\pi$ gives $m\in \mathbb{Z}$, with an angular eigenbasis $\boxed{\Phi =e^{im\varphi }}$. Linear combinations of the degenerate $m,-m$ modes are also possible.

Inside the circular infinite quantum well, put $\rho =r\sqrt{2\mu E/{\hslash }^2}$. The equation becomesThe physical solution must be regular at the origin, behaving as $r^{|m|}$ for $m\ne 0$, or with finite value and zero radial derivative for $m=0$. At the infinite wall impose $R(r=a)=0$. The printed question's boundary phrase uses capital $R=a$; it means the radial position $r=a$, not a value imposed on the radial wavefunction. The singular Bessel function of the second kind is excluded; mere square-integrability of its logarithmic $m=0$ singularity would not put it in the ordinary finite-energy Hamiltonian domain.

For $m=0$, multiply the equation by ${\rho }^2$ and insert $R={\sum }_{k\ge 0}{A}_k{\rho }^k$. The coefficient of $\rho$ gives $A_1=0$, and subsequent coefficients giveAll odd coefficients vanish, $A_2=-A_0/4$, and $A_4=A_0/64$. The regular solution is a multiple of the Bessel function of the first kind $J_0$. Its requested fourth-order estimate isTherefore the first positive zero of this truncation is $\boxed{\rho \approx \sqrt{8}}$, givingThe ground state is in the $m=0$ sector because nonzero angular modes add nonnegative centrifugal kinetic energy to the radial energy form. This quartic truncation of the Bessel function J0 is only a rough local-series estimate, not an exact eigenvalue or a certified variational bound; its double zero warns that the truncation does not reproduce the simple zero of the full function.

For an axisymmetric Newtonian gravitational potential, the vacuum Laplace equation in cylindrical coordinates is $R^{-1}{\partial }_R(R{\partial }_R\Phi )+{\partial }_z^2\Phi =0$. Use separation of variables, $\Phi =X(R)Z(z)$. Dividing by $XZ$ and setting the separated constant to $k^2$, with $k>0$, givesThe first equation has solutions $e^{\pm kz}$; with $u=kR$, the second is the order-zero Bessel differential equation. Regularity at the axis selects the Bessel function of the first kind $J_0(kR)$, rather than the singular Bessel function of the second kind $Y_0$. ThusThere are also zero-separation-constant modes, and other boundary conditions can select other separated families. For a localized reflection-symmetric thin disk, decay away from the disk selects $e^{-k|z|}$ on the respective half-spaces.

Apply the divergence theorem to the Poisson equation for Newtonian gravity across a narrow pillbox enclosing disk area. Writing the volume mass density as $\Sigma (R)\delta (z)$ givesReflection symmetry makes ${\Phi }_z(R,0^{+})=2\pi G\Sigma (R)$. A superposition of the regular vacuum modes therefore satisfiesThe order-zero Hankel transform and its inversion now determine the Hankel representation of thin-disk gravity:The sign is fixed by the positive derivative of an attractive Newtonian gravitational potential above the disk. For ordinary finite disks the representation uses the isolated boundary condition, fixing any extra vacuum field or additive constant. Infinite scale-free disks require suitable regularization of the potential itself.

Equatorial circular orbit balance is $v_c^2=R{\Phi }_R(R,0)$. Using $J_0^{\prime }(u)=-J_1(u)$ in the Hankel representation of thin-disk gravity givesDifferentiation under the integral is justified by convergence for regular models, or by first working off the disk and taking the limiting radial field.

Write $C={\Sigma }_0{R}_0$. For the Mestel disk, the supplied Bessel function integral gives $S(k)=-2\pi GC/k$ for $k>0$. The circular speed is thenMeanwhile the interior mass is $M(R)=2\pi {\int }_0^R(C/s)sds=2\pi CR$. ConsequentlyThe Mestel disk has a flat galaxy rotation curve, despite its nonspherical geometry.

Its infinite total mass prevents setting the Newtonian gravitational potential to zero at infinity. Indeed the integral for $\Phi$ diverges at $k=0$ because $S(k)\sim 1/k$. Subtracting the potential at a reference point removes this additive divergence. As an explicit check, the Mestel disk potential-density pair isIt is harmonic for $z\ne 0$, its vertical derivative jump is $4\pi GC/R$, and $R{\Phi }_R(R,0)=2\pi GC$. Equivalently, an Abel regularization of an oscillatory integral with damping factor $e^{-ak}$ in the radial-field integral gives $R{\int }_0^{\infty }{e}^{-ak}{J}_1(kR)dk=1-a/\sqrt{a^2+R^2}$, which tends to one as $a\downarrow 0$. The force is well-defined although the unreferenced potential is not.

The equality with $GM(R)/R$ is special to the Mestel disk, not a thin-disk shell theorem. The spherical shell theorem normally makes that formula possible, but does not apply to an astrophysical disk. Exterior annuli exert outward radial gravitational acceleration inside their radii, while interior annuli do not act exactly as point masses at the origin. For this particular scale-free surface density, their combined contributions happen to produce the same result as the interior mass expression. Changing the radial surface density or truncating the disk generally destroys the equality: enclosed mass does not determine a disc rotation curve.

For small $x$, use $u=\cosh t$ to write the defining oscillatory integral asThe first integral is $-\mathrm{Ci}(x)$, where the cosine integral satisfies $\mathrm{Ci}(x)={\gamma }_E+\log x+O(x^2)$. Its logarithm follows by integrating the $1/x$ singularity of its derivative; its conventional constant is the Euler--Mascheroni constant. The primary normalization is given by dlmf.nist.gov/6.6.E6 .

The second integrand has an integrable inverse-square-root endpoint singularity and an $O(u^{-3})$ tail. The dominated convergence theorem therefore gives its limitSplitting at $u=1/x$ and using $|1-\cos (xu)|\le \min (x^2{u}^2/2,2)$ also bounds its error by $O(x^2|\log x|)$. ThusIn particular, the leading term is $(2/\pi )\log x$.

For large $x$, the phase $\cosh t$ has its only stationary point at the endpoint $t=0$. Its expansion is $1+t^2/2+O(t^4)$, so the contributing width is $t=O(x^{-1/2})$. The endpoint stationary phase method and the Fresnel integral yieldA smooth cutoff near the endpoint justifies this local replacement; the remaining nonstationary tail can be integrated by parts, with no stationary contribution at infinity. Taking the real part gives the Bessel function of the second kind asymptoticThe additive error formulation remains meaningful near zeros of the leading oscillation. Both limits were derived from the given integral, rather than identifying a special function and merely quoting its asymptotics.

Put $\varphi =R(r)\Theta (\theta )Z(z)$ in the Laplace equation in cylindrical coordinates. Separation of variables first gives ${\Theta }^{\prime \prime }+m^2\Theta =0$; $2\pi$-periodicity forces $m=0,1,2,\dots$, with angular factors $\cos m\theta ,\sin m\theta$ (only the constant for $m=0$). Taking $Z^{\prime \prime }={\kappa }^{2}Z$ givesFor $\kappa >0$ the separated modes are thereforeHere $J_m$ and $Y_m$ are the Bessel function of the first kind and the Bessel function of the second kind. For the opposite sign $Z^{\prime \prime }=-{\kappa }^{2}Z$, the modes instead involve $I_m(\kappa r),K_m(\kappa r)$, the modified Bessel functions, paired with $\cos \kappa z,\sin \kappa z$. At zero separation constant use $Z=C+Dz$ and $R=A{r}^m+B{r}^{-m}$ for $m\ge 1$, or $R=A+B\log r$ for $m=0$. Superpositions over the integer angular orders and appropriate sums or integrals over the separation parameter give the general separated expansion; boundary conditions select its spectrum and coefficients. Regularity at the axis excludes $Y_m,K_m,r^{-m},\log r$ terms.

For the specified decaying side data, bounded separated modes use $e^{-\kappa z}{J}_m(\kappa r)$. A particular solution isAll three denominators are nonzero. Each term satisfies the radial Bessel differential equation, is regular at the axis and bounded for $0\le r\le 1,z\ge 0$, and substitution at $r=1$ gives the stated side values. The modes behave as $r^m$ at the axis, so their apparent angular dependence there causes no singularity.

There is an actual nonuniqueness in the printed problem: no values are specified at $z=0$. If $j_{0,1}$ is the first positive zero of $J_0$, then for every real $a$,is another bounded solution with the same side values. It remains smooth at the axis and even decays as $z\to \infty$. This explicitly proves that side boundary data do not determine a bounded harmonic function in a half-cylinder. Thus the boxed expression supplies a bounded solution; the phrase “the bounded solution” is not justified without an additional base boundary condition. The PDF has $z\ge 0$, whereas the TeX transcription displays $z>0$.

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.

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).