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For constant positive Hubble parameter $\mu$ and a four-thirds polytropic equation of state, a linear cosmological density perturbation obeys $\ddot{\delta }+2\mu \dot{\delta }+{\eta }^2{e}^{-3\mu t}\delta =0$. Put $\delta =e^{-\mu t}y$ and $x=2|\eta |e^{-3\mu t/2}/(3\mu )$. The resulting Bessel differential equation has order $2/3$; use ordinary Bessel functions for ${\eta }^2>0$ and modified Bessel functions for ${\eta }^2<0$. Their small-$x$ expansions give one constant limiting mode and one $e^{-2\mu t}$ mode. The constant limit need not exceed the initial amplitude without an initial-velocity condition.

The Bessel function of the first kind $J_{\nu }$ is the solution of the Bessel differential equation selected by its regular power-series behavior at the origin. For $\mathrm{Re}\nu >-1/2$, it has the Poisson integral representation

For noninteger order, $Y_{\nu }=(\cos (\pi \nu )J_{\nu }-J_{-\nu })/\sin (\pi \nu )$; the integer-order function is defined by a limit. It gives a second independent solution of the Bessel differential equation. At order zero, $Y_0(z)\sim (2/\pi )[\log (z/2)+\gamma ]$ near positive zero and $Y_0(z)\sim \sqrt{2/(\pi z)}\sin (z-\pi /4)$ at large positive argument, where $\gamma$ is the Euler--Mascheroni constant.

An oscillator with frequency $\omega (t)=e^{-\epsilon t}$ leaves the WKB approximation when $|{\omega }_t|/{\omega }^2=\epsilon e^{\epsilon t}$ becomes order one. The shifted variable $T=\epsilon t-\log ({\epsilon }^{-1})$ changes its leading undamped equation to $Y_{TT}+e^{-2T}Y=0$. With $Z=e^{-T}$ this is the order-zero Bessel differential equation. Large-$Z$ Bessel functions match the earlier oscillations, while the small-$Z$ constant and logarithmic solutions become a constant and a linear function of late time.

The two Hankel functions $H_{\nu }^{(1)}=J_{\nu }+i{Y}_{\nu }$ and $H_{\nu }^{(2)}=J_{\nu }-i{Y}_{\nu }$ solve the Bessel differential equation. Their large-argument forms distinguish outgoing and incoming oscillations and have sector-dependent asymptotic expansions.

Write $T=\epsilon t$ and $q(T)=2e^{-T}$. The forcing acts on the fast initial-time scale, so first remove a localized particular solution. Expanding the coefficient there givesA particular choice is $p_1=(t+1/2)e^{-2t}$, henceIts initial value is $1+\epsilon /2$ and its initial derivative is $-2+O({\epsilon }^2)$. The localized forcing determines WKB phase constants; omitting it before imposing the initial conditions would miss the permanent free oscillation.

For the homogeneous part, the slow-time equation is ${\epsilon }^2{u}_{TT}+q^{2}u=0$. Write its WKB approximation as $A(T)[C\cos \Phi +B\sin \Phi ]$, normalizing $A(0)=1$. The leading amplitude transport equation gives $A=e^{T/2}$ and leading phase ${\epsilon }^{-1}{\int }_0^{T}q(s)ds=2(1-e^{-T})/\epsilon$.

To include every $O(\epsilon )$ contribution on bounded slow-time intervals, the first phase correction is needed. If the local phase wavenumber is $q+{\epsilon }^2{q}_2$, the real part of the amplitude-phase equation gives $q_2=A_{TT}/(2qA)=e^T/16$. The WKB phase correction for a slowly varying oscillator, anchored to zero at the initial time, is thereforeThere is no first-order bulk amplitude correction in this normalization. At $t=0$, the homogeneous initial value must be $-\epsilon /2+O({\epsilon }^2)$ and its initial derivative $1+O({\epsilon }^2)$. Since $A_t^{\prime }(0)=\epsilon /2$ and ${\Phi }_t^{\prime }(0)=2+O({\epsilon }^2)$, these give $C=-\epsilon /2+O({\epsilon }^2)$ and $B=1/2+O({\epsilon }^2)$. The requested forced approximation isfor $0\le \epsilon t\le T_0$ with fixed $T_0$. The approximate initial value is exactly one through this order and its derivative is $-1+O({\epsilon }^2)$. The particular-solution residual is uniformly localized and $O({\epsilon }^2)$; the corrected homogeneous residual is $O({\epsilon }^3)$ on the long interval, producing $O({\epsilon }^2)$ solution error after integration over times of order ${\epsilon }^{-1}$.

The WKB approximation requires the relative frequency-change parameterIt therefore fails when $\epsilon e^{\epsilon t}=O(1)$, even though the coefficient has no finite-time zero. This is the Bessel transition for an exponentially decaying oscillator. Introduce the shifted slow time and amplitudeThe exact rescaled equation isFor fixed $\tau$ its right side is smaller than every algebraic power of $\epsilon$. The leading transition problem is consequentlyWith $Z=2e^{-\tau }$ this becomes $Z^2{U}_{0,ZZ}+Z{U}_{0,Z}+Z^2{U}_0=0$, the order-zero Bessel differential equation. This identifies the governing equation without solving it.

The amplitude-phase series also supplies overlap data while $\epsilon e^T\ll 1$, although its bounded-$T$ absolute error estimate is not uniform up to the transition. For matching, combine the earlier sine and cosine terms into $\frac{1}{2}\sin (\Phi -\epsilon )$ through first order. In the overlap $\epsilon \ll e^{\tau }\ll 1$, their phase isto the retained order. Thus amplitude and phase matching as $\tau \to -\infty$ requireThese are oscillatory amplitude-phase conditions, rather than pointwise ratio statements at zeros. The smaller $e^{\tau }/16$ phase term records the first overlap correction if required. The rapidly varying matching constant ${\alpha }_{\epsilon }$ must be retained as data; there is no single epsilon-independent limiting phase. The two matching conditions determine the leading transition solution, which is deliberately left unsolved.

Put $B=E{I}_2$, the filament bending modulus in the easier bending direction; $I_2$ is the relevant second moment of area. Take filament tension as positive in extension. A cut at height $z$ supports the weight of the material above it, so its axial force is compressive:The small transverse displacement induces a transverse tension force density $(T{X}_z{)}_z$, while bending produces $-B{X}_{zzzz}$. Their balance givesThe same result follows from elastic energy and gravitational energy. To quadratic order, local inextensibility shortens the height of the material point labelled $z$ by $\frac{1}{2}{\int }_0^z{X}_z(s{)}^{2}ds$. Integrating its weight and interchanging the integrals gives the second-order potential energyTwo integration by parts steps in its first variation giveThe clamped boundary conditions fix displacement and slope at the base, and the unconstrained tip must have zero bending moment and transverse end force. Thus the complete four conditions areSince the filament tension vanishes at the tip, the last condition is $X_{zzz}(L)=0$. Integrate the ordinary differential equation once: $B{X}_{zzz}-T{X}_z$ is constant, and its tip value makes that constant zero. For the slope $u=X_z$, it follows thatThe other tip condition $u_{zz}(L)=0$ already follows from this equation for a regular solution; it is not a third independent condition on the second-order slope problem. Finally reconstruct $X(z)={\int }_0^{z}u(s)ds$ to impose the base position.

Set $x=L-z$, $\alpha =\lambda g/B$, and $\eta =(2/3)\sqrt{\alpha }{x}^{3/2}$. Then $d\eta /dx=3\eta /(2x)$ and $d^2\eta /d{x}^2=3\eta /(4x^2)$, so $u_{xx}+\alpha xu=0$ becomesSubstituting $u={\eta }^{1/3}F(\eta )$ givesThis is the Bessel differential equation of order $1/3$, whose two independent Bessel functions of the first kind giveThe tip is $x=\eta =0$. To decide which branch is admissible, retain both leading behaviours, not just the fact that $u$ stays finite:Since ${\eta }^2\propto x^3$ and ${\eta }^{2/3}\propto x$, the first branch has zero tip derivative, while the second has a nonzero tip derivative. The moment-free condition therefore sets $b=0$, and it does not require the tip slope itself to vanish. The base condition now requiresLet $j_{-1/3,1}$ be the first positive zero. The Bessel threshold for self-buckling of a vertical rod isTo identify this as the first loss of stability, write the quadratic energy in terms of the slope as $\frac{1}{2}[B\int u_z^{2}dz-\lambda g\int (L-z)u^{2}dz]$. On functions with $u(0)=0$, its critical coefficient is the minimum Rayleigh quotient $\int u_z^{2}dz/\int (L-z)u^{2}dz$; variation supplies the moment-free condition at the tip and the same Sturm-Liouville problem. The first zero gives the lowest eigenvalue, so the straight state is stable below this load, neutral at it, and unstable above it, with the easier $I_2$ direction buckling first.

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.

Take the logarithmic derivative of the proposed asymptotic expansion. Through order $z^{-1}$, $u^{\prime }/u=\lambda +\mu /z+O(z^{-2})$ and $u^{\prime \prime }/u={\lambda }^2+2\lambda \mu /z+O(z^{-2})$. Matching the constant and inverse-power terms givesThe second formula assumes a simple root, $2\lambda +f_0\ne 0$. At a repeated root the ansatz can fail: the next equation first requires $f_1\lambda +g_1=0$, and if that holds higher orders determine the remaining behaviour.

For the Bessel differential equation, $\lambda =\pm i$ and $\mu =-1/2$. Set $u=e^{\lambda z}{z}^{-1/2}v(z)$. Direct substitution leaves $v^{\prime \prime }+2\lambda v^{\prime }+(1/4-{\nu }^2)z^{-2}v=0$. With $v=1+a/z+O(z^{-2})$, matching $z^{-2}$ gives $a=(1/4-{\nu }^2)/(2\lambda )$. ThusTwo constant multiples of the Hankel functions normalize their leading coefficients to one and realize these solutions in their appropriate sectors.

For generic $\nu$, these are sectorial expansions, not unchanged expansions for a single branch over arbitrarily continued $\arg z$. The factors $z^{-1/2}$ require a branch, and crossing sector boundaries can introduce the other exponential by Stokes phenomenon and analytic continuation. More precisely, the standard normalized first Hankel solution has its expansion for $-\pi +\delta \le \arg z\le 2\pi -\delta$, and the second for $-2\pi +\delta \le \arg z\le \pi -\delta$, with any fixed positive $\delta$ inside these sectors. In particular both are valid in the common slit sector $-\pi <\arg z<\pi$, away from its boundaries. These sector statements are recorded in NIST's Hankel asymptotics. Exceptional half-integer orders have terminating elementary expressions, so a blanket assertion of nonzero Stokes mixing for every order would be too strong.

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.

Fora finite point $z_0$ is a regular singular point when $(z-z_0)P(z)$ and $(z-z_0{)}^{2}Q(z)$ extend holomorphically to $z_0$. A singularity at infinity is classified by applying this criterion after the change of variable $w=1/z$.

For the displayed Bessel differential equation,At $z=0$, the functions $zP(z)=1$ and $z^{2}Q(z)=z^2-1/4$ are holomorphic, so zero is regular singular. To inspect infinity, write $Y(w)=y(1/w)$. The equation becomesHere $w^{2}Q(w)=w^{-2}-1/4$ is not holomorphic at zero, so infinity is an irregular singular point. Thus

Set $y=z^{-1/2}f$. Direct differentiation makes every lower-order term cancel:Hence $f^{\prime \prime }+f=0$, and on any domain carrying a branch of $\sqrt{z}$ two linearly independent solutions areNear the regular singular point, these behave as $z^{-1/2}$ and $z^{1/2}$, the two Frobenius exponents. At infinity their combinations are $z^{-1/2}{e}^{\pm iz}$; after $w=1/z$ these contain $e^{\pm i/w}$, the exponential behavior characteristic of an irregular singular point.

Since $x=r\cos \theta$, the function on the left of the defining expansion is $e^{ix}$. Thereforewhich is the two-dimensional Helmholtz equation.

In polar coordinates,Substitute the defining Fourier cosine series and equate the coefficient of each $\cos (n\theta )$. Its angular second derivative contributes $-n^2$, so every Bessel function $J_n$ satisfiesor equivalently the Bessel differential equation

Expand the exponential through cubic order and useThis givesComparing with $J_0+2{\sum }_{n\ge 1}{i}^n{J}_n\cos (n\theta )$ yields

LetTheir Bessel equations in self-adjoint form areMultiply the first by $v$, the second by $u$, subtract, and integrate. The left side becomes a boundary term:Regularity at zero makes the lower boundary term vanish, whileTherefore

If $\alpha ={\gamma }_k$ and $\beta ={\gamma }_{\ell }$ with $k\ne \ell$, both endpoint values of $J_0$ vanish, so the integral is zero. This is the weighted orthogonality of distinct eigenfunctions in a Sturm-Liouville problem. For the norm, hold $\alpha ={\gamma }_k$ and let $\beta \to {\gamma }_k$. Sincethe quotient tends to

The collapse is perpendicular to the initial field, so magnetic flux freezing preserves the mass-to-flux ratio of each material flux tube:With no toroidal field, radial magnetostatic equilibrium isDefine the effective polytropic constantThen gas and magnetic pressure combine as $P+B_z^2/(8\pi )=K_{\mathrm{eff}}{\rho }^2$. Dividing equilibrium by $\rho$, differentiating, and using the cylindrical Poisson equationgivesPut $a^2=K_{\mathrm{eff}}/(2\pi G)$. This is the order-zero Bessel differential equation, and regularity on the axis together with $\rho (0)={\rho }_1$ yields

Take $z$ upward along the straight rod. The part above height $z$ has weight $\lambda g(L-z)$, so, with tensile force positive, the internal tension is compressive:For a small transverse displacement $X(z)$, the quadratic bending and gravitational energies areThe Euler-Lagrange equation for this functional iswhich is exactly

Clamping at the bottom fixes displacement and slope:At the free upper end, bending moment and transverse force vanish:Since $\sigma (L)=0$, the four boundary conditions are

Set $u=X^{\prime }$. Integrating the field equation once and using the free-end shear condition givesor, with $s=L-z$,Introduce the dimensionless similarity coordinateand write $u={\eta }^{1/3}F(\eta )$. Direct substitution reduces the equation toThis is the Bessel differential equation of order $1/3$, so