This shows that if F(z) be any single valued monogenic function which is doubly periodic and of meromorphic character, then F(z + t) is an algebraic function of F(z) and F(t). Conversely any single valued monogenic function of meromorphic character, F(z), which is such that F(z + t) is an algebraic function of F(z) and F(t), can be shown to be a doubly periodic function, or a function obtained from such by degeneration (in virtue of special relations connecting the fundamental constants).
The functions ƒ(z), φ(z) above are usually denoted by ℜ(z), ℜ′(z); further the fundamental differential equation is usually written
(ℜ′z)² = 4(ℜz)³ − g2ℜz − g3,
and the roots of the cubic on the right are denoted by e1, e2, e3; for the odd function, ℜ′z, we have, for the congruent arguments −½ωand ½ω, ℜ′ (½ω) = −ℜ′ (−½ω) = −ℜ′ (½ω), and hence ℜ′ (½ω) = 0; hence we can take e1 = ℜ (½ω), e2 = ℜ (½ω + ½ω′), e3 = ℜ (½ω). It can then be proved that [ℜ(z) − e1] [ℜ (z + ½ω) − e1] = (e1 − e2) (e1 − e3), with similar equations for the other half periods. Consider more particularly the function ℜ(z) − e1; like ℜ(z) it has a pole of the second order at z = 0, its expansion in its neighbourhood being of the form z−2 (1 − e1z2 + Az4 + ...); having no other pole, it has therefore either two zeros, or a double zero in a period parallelogram (ω, ω′). In fact near its zero ½ω its expansion is (x − ½ω) ℜ′ (½ω) + ½(z − ½ω)² ℜ″ (½ω) + ...; we have seen that ℜ′ (½ω) = 0; thus it has a zero of the second order wherever it vanishes. Thus it appears that the square root [ℜ(z) − e1]1/2, if we attach a definite sign to it for some particular value of z, is a single valued function of z; for it can at most have two values, and the only small circuits in the plane which could lead to an interchange of these values are those about either a pole or a zero, neither of which, as we have seen, has this effect; the function is therefore single valued for any circuit. Denoting the function, for a moment, by ƒ1(z), we have ƒ1(z + ω) = ±ƒ1(z), ƒ1(z + ω′) = ±ƒ1(z); it can be seen by considerations of continuity that the right sign in either of these equations does not vary with z; not both these signs can be positive, since the function has only one pole, of the first order, in a parallelogram (ω, ω′); from the expansion of ƒ1(z) about z = 0, namely z− 1 (1 − ½e1z² + ...), it follows that ƒ1(z) is an odd function, and hence ƒ1 (−½ω′) = −ƒ1 (½ω′), which is not zero since [ƒ1 (½ω′)]² = e3 − e1, so that we have ƒ1 (z + ω′) = −ƒ1(z); an equation f1(z + ω) = −ƒ1(z) would then give ƒ1(z + ω + ω′) = ƒ1(z), and hence ƒ1(½ω + ½ω′) = ƒ1(−½ω − ½ω′), of which the latter is −ƒ1(½ω + ½ω′); this would give ƒ1(½ω + ½ω′) = 0, while [ƒ1(½ω + ½ω′)]² = e2 − e1. We thus infer that ƒ1(z + ω) = ƒ1(z), ƒ1(z + ω′) = −ƒ1(z), ƒ1(z + ω + ω′) = −ƒ1(z). The function ƒ1(z) is thus doubly periodic with the periods ω and 2ω′; in a parallelogram of which two sides are ω and 2ω′ it has poles at z = 0, z = ω′ each of the first order, and zeros of the first order at z = ½ω, z = ½ω + ω′; it is thus a doubly periodic function of the second order with two different poles of the first order in its parallelogram (ω, 2ω′). We may similarly consider the functions ƒ2(z) = [ℜ(z) − e2]1/2, ƒ3(z) = [ℜ(z) − e3]1/2; they give
| ƒ2(z + ω + ω′) = ƒ2(z), ƒ2(z + ω) = −ƒ2(z), ƒ2(z + ω′) = −ƒ2(z), | ƒ3(z + ω′) = ƒ3z, ƒ3(z + ω) = −ƒ3(z), ƒ3(z + ω + ω′) = −ƒ3(z). |
Taking u = z (e1 − e3)1/2, with a definite determination of the constant (e1 − e3)1/2, it is usual, taking the preliminary signs so that for z = 0 each of zƒ1(z), zƒ2(z), zƒ3(z) is equal to +1, to put
| sn(u) = | (e1 − e3)1/2 | , cn(u) = | ƒ1(z) | , dn(u) = | f2(z) | , |
| ƒ3(z) | ƒ3(z) | ƒ3(z) |
| k² = (e2 − e3) / (e1 − e3), K = ½ω (e1 − e3)1/2, iK′ = ½ω′ (e1 − e3)1/2; |
thus sn(u) is an odd doubly periodic function of the second order with the periods 4K, 2iK, having poles of the first order at u = iK′, u = 2K + iK′, and zeros of the first order at u = 0, u = 2K; similarly cn(u), dn(u) are even doubly periodic functions whose periods can be written down, and sn²(u) + cn²(u) = 1, k²sn²(u) + dn²(u) = 1; if x = sn(u) we at once find, from the relations given here, that
| du | = [(1 − x²) (1 − k²x²)]−1/2; |
| dx |