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For an axisymmetric reflection-symmetric thin disk, the Laplace equation in cylindrical coordinates separates into regular radial Bessel functions and vertical decaying exponentials. The divergence theorem imposes ${\Phi }_z^{+}-{\Phi }_z^{-}=4\pi G\Sigma$. Hankel transform inversion gives $S(k)=-2\pi G{\int }_0^{\infty }\Sigma (R)J_0(kR)RdR$. Its radial derivative gives the galaxy rotation curve. Infinite scale-free models may require subtracting an additive divergent constant.

The separated disk modes. Away from the plane of a razor-thin disk, there is no volume source, so the Newtonian gravitational potential obeys the Laplace equation in cylindrical coordinates. Separation of variables givesHere $k>0$ is a radial wavenumber, with units of inverse length. Regularity on the axis selects the Bessel function of the first kind $J_0(kR)$ rather than the singular second solution. Decay away from the disk, continuity across it and reflection symmetry select $Z=e^{-k|z|}$. These conditions apply to each nonzero-wavenumber mode; a uniform sheet is a separate zero-mode problem.

Integrate the Poisson equation for Newtonian gravity across the disk to obtain the jump conditionFor the unit-amplitude separated mode, the jump is $-2k{J}_0(kR)$. ThusAn individual basis density is signed; positive physical surface densities are built by superposition.

An arbitrary axisymmetric surface density. Define its order-zero Hankel transform byMatching the coefficients of the mode densities gives the Hankel representation of thin-disk gravity,The boundary conditions exclude added external fields. For sufficiently decaying surface density, this is the isolated-disk potential; for infinite scale-free disks, only potential differences or regularized forces need be finite.

The Mestel disk. Put $C={\Sigma }_0{R}_0$. The Bessel function integral ${\int }_0^{\infty }{J}_0(k{R}^{\prime })d{R}^{\prime }=1/k$, understood with a decaying regulator, gives $\hat{\Sigma }=C/k$. Differentiate the potential before removing the regulator and use $J_0^{\prime }=-J_1$:The last integral is $1/R$, since ${\int }_0^{\infty }{J}_1(u)du=[-J_0(u){]}_0^{\infty }=1$. Therefore the circular speed and enclosed mass areConsequently $v_c^2=GM(R)/R$. This equality follows from the specific Mestel disk calculation; the spherical shell theorem cannot be applied to general disks. Its total mass and absolute potential referenced to infinity diverge. A convenient finite midplane reference is $\Phi (R,0)=v_0^2\log (R/R_0)$.

The rotating distribution function. Use specific energy $E=(v_R^2+v_{\varphi }^2)/2+\Phi$ and $L_z=R{v}_{\varphi }$. For the printed positive exponential, take $\epsilon =-E$ and $s:=\sigma >0$; $s$ has units of squared velocity. If instead $\epsilon$ means ordinary total energy and $\sigma >0$, the velocity integral diverges, so that literal convention cannot define the requested galactic distribution function. Equivalently one can retain ordinary energy and choose the printed $\sigma$ negative.

With the binding-energy convention, write the amplitude as $A$ to distinguish it from the function:Integration over both radial velocities and positive azimuthal velocities givesThe half-line integral follows by setting $u=v_{\varphi }^2/(2s)$ in the defining Gamma function integral and requires $q>-1$. Matching its radial power and normalization to the Mestel disk yields the rotating Mestel disk distribution function parametersSince $v_0^2,s>0$, the required integrability condition is automatic. The amplitude uses the stated potential reference; changing the additive potential constant changes $A$ without changing the physical distribution.

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.

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 $z>0$.

For an axisymmetric separated amplitude $\hat{\varphi }(r,z)=R(r)Z(z)$, Laplace equation in cylindrical coordinates becomesThe radial equation isIts solution regular at the axis is the Bessel function $J_0(kr)$. The vertical equation is $Z^{\prime \prime }-k^{2}Z=0$, and the bottom condition $Z^{\prime }(-h)=0$ selectsup to an irrelevant constant multiplier. Finally, the sidewall condition givesTherefore the allowed positive wavenumbers areand