d²y/dt² + (2t−1 − at−2) dy/dt + bt−4 y = 0,
which is not regular about t = 0 unless 2 − at−1 and bt−2, that is, unless ax and bx² are finite at x = ∞; which we thus assume; putting y = tr(1 + A1t + ... ), we find for the index equation at x = ∞ the equation r(r − 1) + r(2 − ax)0 + (bx²)0 = 0. If there be Equation of the second order. finite singular points at ξ1, ... ξm, where we assume m > 1, the cases m = 0, m = 1 being easily dealt with, and if φ(x) = (x − ξ1) ... (x − ξm), we must have a·φ(x) and b·[φ(x)]² finite for all finite values of x, equal say to the respective polynomials ψ(x) and θ(x), of which by the conditions at x = ∞ the highest respective orders possible are m − 1 and 2(m − 1). The index equation at x = ξ1 is r(r − 1) + rψ(ξ1) / φ′ (ξ1) + θ(ξ)1 / [φ′(ξ1)]² = 0, and if α1, β1 be its roots, we have α1 + β1 = 1 − ψ(ξ1) / φ′ (ξ1) and α1β1 = θ(ξ)1 / [φ′(ξ1)]². Thus by an elementary theorem of algebra, the sum Σ(1 − αi − βi) / (x − ξi), extended to the m finite singular points, is equal to ψ(x) / φ(x), and the sum Σ(1 − αi − βi) is equal to the ratio of the coefficients of the highest powers of x in ψ(x) and φ(x), and therefore equal to 1 + α + β, where α, β are the indices at x = ∞. Further, if (x, 1)m−2 denote the integral part of the quotient θ(x) / φ(x), we have Σ αiβiφ′ (ξi) / (x = ξi) equal to −(x, 1)m−2 + θ(x)/φ(x), and the coefficient of xm−2 in (x, 1)m−2 is αβ. Thus the differential equation has the form
y″ + y′Σ (1 − αi − βi) / (x − ξi) + y[(x, 1)m-2 + Σ αiβiφ′(ξi) / (x − ξi)]/φ(x) = 0.
If, however, we make a change in the dependent variable, putting y = (x − ξ1)α1 ... (x − ξm)α mη, it is easy to see that the equation changes into one having the same singular points about each of which it is regular, and that the indices at x = ξi become 0 and βi − αi, which we shall denote by λi, for (x − ξi)αj can be developed in positive integral powers of x − ξi about x = ξi; by this transformation the indices at x = ∞ are changed to
α + α1 + ... + αm, β + β1 + ... + βm
which we shall denote by λ, μ. If we suppose this change to have been introduced, and still denote the independent variable by y, the equation has the form
y″ + y′Σ (1 − λi) / (x − ξi) + y(x, 1)m−2 / φ(x) = 0,
while λ + μ + λ1 + ... + λm = m − 1. Conversely, it is easy to verify that if λμ be the coefficient of xm−2 in (x, 1)m−2, this equation has the specified singular points and indices whatever be the other coefficients in (x, 1)m−2.
Thus we see that (beside the cases m = 0, m = 1) the “Fuchsian equation” of the second order with two finite singular points is distinguished by the fact that it has a definite form when the singular points and the indices are assigned. Hypergeometric equation. In that case, putting (x − ξ1) / (x − ξ2) = t / (t − 1), the singular points are transformed to 0, 1, ∞, and, as is clear, without change of indices. Still denoting the independent variable by x, the equation then has the form
x(1 − x)y″ + y′[1 − λ1 − x(1 + λ + μ)] − λμy = 0,