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Associated Legendre functions solve $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }+[\nu (\nu +1)-{\mu }^2/(1-x^2)]y=0$. For integers $\ell \ge 0$ and $0\le m\le \ell$, the regular solution is $P_{\ell }^m(x)=(-1{)}^m(1-x^2{)}^{m/2}(d/dx{)}^m{P}_{\ell }(x)$, using a Legendre polynomial. It enters normalized spherical harmonics. Despite the traditional term associated Legendre polynomial, the factor $(1-x^2{)}^{m/2}$ is not a polynomial when $m$ is odd. Negative integer orders satisfy $P_{\ell }^{-m}=(-1{)}^m(\ell -m)!P_{\ell }^m/(\ell +m)!$.

For density proportional to ${\mathrm{sech}}^2(z/H)$, the magnetic vertical equation $F^{\prime \prime }+k^2{\mathrm{sech}}^2(z/H)F=0$ becomes the Legendre differential equation under $\xi =\tanh (z/H)$. Bounded solutions at both surfaces are Legendre polynomials, with $k_n=\sqrt{n(n+1)}/H$. The $n=0$ constant function has no magnetically coupled finite wavenumber. The smallest MRI-relevant eigenvalue is $k_1^2=2/H^2$, which makes sufficiently strong magnetic tension stabilize all allowed modes.

An axisymmetric harmonic potential outside a sphere of radius $r_0$, with far field $v_{0}r\cos \theta$ and zero radial derivative on the sphere, isThe regular spherical expansion in Legendre polynomials fixes all coefficients. A spatial constant remains arbitrary if only the far-field gradient is specified, and is fixed by prescribing the potential difference at infinity.

The equation $(1-x^2)y\prime \prime -2xy\prime +\ell (\ell +1)y=0$ has polynomial solutions for nonnegative integer $\ell$.

For a thin axisymmetric astrophysical disk, angular integration of the Newtonian kernel eliminates odd Legendre polynomials. Define $I_n={\int }_0^R\Sigma (s)s^{n+1}ds$ and $J_n={\int }_R^{\infty }\Sigma (s)s^{-n}ds$. The even coefficient is ${\alpha }_n=\pi [n!/(2^n((n/2)!{)}^2){]}^2$. In differentiating the paired inner and outer contributions, their moving-boundary terms cancel. The resulting galaxy rotation curve includes an inward inner-ring contribution and an outward exterior-ring contribution. For smooth surface density, the force at the disk is interpreted with a symmetric Cauchy principal value or a vanishing-thickness regularization.

An $s$-node Lobatto quadrature rule includes both endpoints and is exact for polynomials of degree at most $2s-3$. On $[-1,1]$, its interior nodes are the zeros of the derivative of the degree-$(s-1)$ Legendre polynomial. The three-node rule on $[0,1]$ has nodes $0,1/2,1$ and weights $1/6,2/3,1/6$. Integrating the interpolation basis from zero to each node gives the three-stage Lobatto IIIA method.

Substitute $y(x)={\sum }_{j\ge 0}{c}_j{x}^j$ into the Legendre differential equation. Matching the coefficient of $x^j$ givesChoose the parity of $n$: start with nonzero $c_0$ for even $n$, or nonzero $c_1$ for odd $n$, and set the other parity to zero. The numerator is nonzero for every preceding index of that parity below $n$, so $c_n\ne 0$, but at $j=n$ it vanishes. All higher coefficients are zero. This constructs a nonzero polynomial solution of degree exactly $n$.

Its value at $x=1$ cannot be zero. If it had a zero there of multiplicity $k\ge 1$, write $y=(x-1{)}^{k}R$ with $R(1)\ne 0$. The lowest-order term of $(1-x^2)y^{\prime \prime }-2x{y}^{\prime }$ would be $-2k^{2}R(1)(x-1{)}^{k-1}$, whereas $n(n+1)y$ starts at order $k$. The differential equation could not hold. Hence the polynomial can be scaled to satisfy $P_n(1)=1$.

The resulting Legendre polynomials beginEach satisfies both the differential equation and the specified normalization.

Linearity of the integral in each argument gives a bilinear form, and $fg=gf$ gives symmetry. Also $(f,f)={\int }_{-1}^{1}f(x{)}^{2}dx\ge 0$. If a polynomial $f$ is nonzero somewhere on this interval, continuity makes $f^2$ strictly positive on a smaller interval, so the integral is positive. A polynomial vanishing on the whole interval is the zero polynomial. Thus the form is positive-definite and is an inner product.

Apply the Gram-Schmidt process to $1,x,x^2$. Odd functions have zero integral, so $x$ is already orthogonal to $1$, and the third orthogonal polynomial is $x^2-1/3$. Their squared norms areTherefore an orthonormal basis isThese are the first three normalized Legendre polynomials. Their distinct degrees also prove that they span the three-dimensional vector space.

A small direction-dependent temperature change $T=\bar{T}(1+\Theta )$ shifts a thermal spectrum according to $\bar{f}(ϵ/(1+\Theta ))=\bar{f}(ϵ)-\Theta ϵ{\bar{f}}^{\prime }(ϵ)$ at first order. Hence $\boxed{\Theta =\delta T/\bar{T}}$ is the angular photon temperature perturbation. For a general spectrum the same ansatz describes an energy-independent brightness dilation; a literal thermodynamic temperature interpretation additionally assumes a thermal shape without spectral distortions.

Use $E=ϵ/a$ in the tetrad stress-energy tensor. The background photon energy density isIntegration by parts gives $-\int dϵ{ϵ}^4{\bar{f}}^{\prime }=4\int dϵ{ϵ}^3\bar{f}$, assuming the boundary term vanishes. Thus each directional energy perturbation is four times its temperature perturbation. Writing $⟨\cdot {⟩}_{\Omega }=(4\pi {)}^{-1}\int d\Omega$, we haveIn the paper's convention for photon angular temperature moments, there is no $(2\ell +1)$ factor multiplying ${\Theta }_{\ell }$. Legendre polynomial orthogonality gives $⟨\Theta {⟩}_{\Omega }={\Theta }_0$ and $⟨e^i\Theta {⟩}_{\Omega }=-i{\hat{k}}^i{\Theta }_1/3$. Comparing with the stated density and velocity conventions yieldsFor completeness, $⟨e^i{e}^j{P}_2(\hat{\mathbf{k}}\cdot \mathbf{e}){⟩}_{\Omega }=({\hat{k}}^i{\hat{k}}^j-{\delta }^{ij}/3)/5$. The $(-i{)}^2$ quadrupole phase then gives $\boxed{{\Pi }_{\gamma }=-3{\Theta }_2/5}$. These moment relations are photon angular temperature moments and keep the anisotropic stress convention consistent with the subsequent hierarchy.

Interpret the printed coordinate notation as the square corners $1+i$, $1-i$, $-1-i$, $-1+i$. This listed order is clockwise, contrary to the counterclockwise convention in (ii). Keep the printed first side directed from top to bottom. Then $m_1=1$, $h_1=-i$, and $z(s)=1-is$. Put $g_1(s)=u(1,-s)$ and $q_1(s)=u_x(1,-s)$, where $q_1$ is the outward normal derivative on the right side. The pullback formula, without any orientation shortcut, givesIndeed $h_1{u}_z-{\bar{h}}_1{u}_{\bar{z}}=-i(u_z+u_{\bar{z}})=-i{u}_x$, and $i\beta (\lambda h_1+{\bar{h}}_1/\lambda )=\beta (\lambda -{\lambda }^{-1})$. If one reverses the side to match the counterclockwise convention, its parameter is $1+is$ and its integrand is $-W_1(-s,\lambda )$. Reversing every side multiplies the global relation by minus one, leaving its zero value unchanged.

The unheaded numerical reconstruction request. The four unknown Neumann boundary conditions are four functions on the sides, not four scalar values. The polygonal global relation is linear in their outward normal derivatives, and its remaining terms depend only on the prescribed Dirichlet boundary data.

Choose a finite approximation on each side, for example an expansion of $q_j(s)$ in Legendre polynomials or piecewise polynomials. Substitute those expansions into the consistently oriented global relation. At chosen nonzero spectral parameters for a linear boundary value problem, integrate the exponential kernels against each basis function to assemble a complex linear system; the known right-hand side is obtained by integrating the given Dirichlet boundary data. Use enough independent samples to resolve all side coefficients, and preferably oversample. The conjugate global relations for the modified Helmholtz equation provide useful companion tests; for complex data use the corresponding independent adjoint relation rather than assuming the traces real.

Solve the scaled system by least-squares solution using a stable factorization such as a singular value decomposition. Sampling directions should probe all sides, and exponential row scaling avoids overflow and poor conditioning. Refine the side approximation and spectral samples until the recovered traces and unused global relation residuals stabilize. Corner incompatibilities or limited corner regularity call for mesh refinement or enriched endpoint basis functions. This realizes a numerical Dirichlet-to-Neumann map without first discretizing the whole interior.

The underlying Dirichlet problem is uniquely solvable in the usual trace class for $\Delta u-4{\beta }^{2}u=0$ with $\beta >0$: the homogeneous problem has ${\int }_{\Omega }(|\nabla u{|}^2+4{\beta }^2|u{|}^2)dA=0$. This supports the boundary reconstruction, although uniqueness of the continuous problem alone does not guarantee that an arbitrary finite set of spectral samples is well conditioned.

For a spherical system, use the spherical shell theorem with shell mass $dm=4\pi \rho (s)s^{2}ds$. Inner shells contribute $-Gdm/r$ to the gravitational potential, while an outer shell contributes its constant interior potential $-Gdm/s$. With $\Phi (\infty )=0$, assuming the required integrals converge,On differentiation, the two terms containing $\rho (r)r$ cancel. Hence ${\Phi }^{\prime }=GM(r)/r^2$, where $M(r)=4\pi {\int }_0^r\rho (s)s^{2}ds$. Radial balance for a circular orbit gives $v_c^2=r{\Phi }^{\prime }=GM(r)/r$.

For a razor-thin axisymmetric astrophysical disk, the element of mass is $dm=\Sigma (s)sdsd{\varphi }^{\prime }$, where $\Sigma$ is the surface density. Superposing Newtonian gravitational potentials therefore givesAxisymmetry permits $\varphi =0$. Unlike the spherical case, an exterior annulus generally exerts a radial force: enclosed mass does not determine a disc rotation curve.

For the Legendre expansion of thin-disk gravity, split the radial integral at $s=R$. In the inner part the kernel expands in $s/R$, while in the outer part it expands in $R/s$. The angular integrals of odd Legendre polynomials vanish, and those of even degree $n$ equal $2{\alpha }_n$, whereIntroduce $I_n(R)={\int }_0^R\Sigma (s)s^{n+1}ds$ and $J_n(R)={\int }_R^{\infty }\Sigma (s)s^{-n}ds$. The gravitational potential becomesWhen differentiating each bracket, the moving-limit terms cancel: $R^{-(n+1)}{I}_n^{\prime }=\Sigma (R)$ and $R^n{J}_n^{\prime }=-\Sigma (R)$. Since $v_c^2=R{\Phi }^{\prime }$, this givesThe zeroth term has ${\alpha }_0=\pi$ and $M(R)=2\pi I_0$. Separating it provesThe inner correction is inward, while the exterior correction is outward. For a smooth surface density, the original potential singularity at coincident points is integrable. Its radial force is understood through a symmetric Cauchy principal value or a vanishing-thickness regularization; the paired interior and exterior terms above retain the cancellation at $s=R$. This avoids treating the two singular local force contributions separately.

For an exponential galactic disk, write $\Sigma (s)={\Sigma }_0{e}^{-s/R_d}$, so $M_{\infty }=2\pi {\Sigma }_0{R}_d^2$ and $M(R)=M_{\infty }[1-(1+R/R_d)e^{-R/R_d}]$. At large $R$, the missing mass and exterior-ring terms are exponentially small. The leading nonspherical interior term is $k=1$, with ${\alpha }_2=\pi /4$ and $I_2(\infty )=6{\Sigma }_0{R}_d^4$. Consequently the exponential-disk Keplerian asymptotic isThe positive leading correction shows that the rotation curve approaches the Keplerian limit from above. The finite-order large-radius expansion is sufficient here; an infinite moment expansion need not converge for a disk extending to infinity.

A useful special example is the Mestel disk, with surface density $\Sigma (s)=C/s$ for $s>0$, $C>0$. It has $M(R)=2\pi CR$. For every positive even $n$,All the nonspherical corrections cancel, givingThus the Mestel disk has a perfectly flat galaxy rotation curve.

This example has infinite total mass and a singular central surface density. The absolute gravitational potential cannot be set to zero at infinity, but the radial force exists as a limit of disks with increasing outer cutoff. Potential differences are $\Phi (R)-\Phi (R_0)=v_M^2\log (R/R_0)$. The earlier potential integral is therefore interpreted up to a radius-independent divergent constant for this example; the force calculation remains valid. Truncating the Mestel disk gives a more physical finite system but changes the exact flat curve near its edges.