A Meromorphic Function

A question in differential geometry, concerning a surface of constant negative curvature, led to the differential equation θ΄ + xθ΄΄ = cos(θ). Although the original problem involved only real variables, curiosity led me to use the computer to graph this equation in the complex plane. I discovered it has ramified singularities, but they appeared to behave very regularly, changing θ by ± 4π when circled. This suggested replacing θ by either tan(θ/4) or exp(iθ/2). The latter proved more convenient, although they are basically equivalent a since linear fractional transformation changes one to the other.

Letting w = exp(iθ/2), the original differential equation becomes zww΄΄ + ww΄ − zw΄² = i/4 + iw⁴/4. It is convenient to generalize this to zww΄΄ + ww΄ − zw΄² = a − bw⁴. Note that u(z) = rw(z) satisfies the same type of equation with r²a and r⁻²b replacing a and b. Using u(z) = w(rz) again gives the same type equation, with a and b multiplied by r. So a and b can be transformed to any non-zero pair. Note, by the way, that linear functions satisfy the equation with b = 0, and their inverses with a = 0.

Because of the factor z in the first term, solutions generally have nasty singularities at z = 0. However, for any non-zero initial value there is a unique solution. One would expect to have singularities also where w becomes zero, because of the w in the first term, as well as where w goes to infinity. However, it turns out that calculations indicate the w΄΄ is zero whenever w is zero, suggesting introduction of a variable v = w΄΄/w. It turns out that v΄ = (4ww΄ – 2v)/z, so the division by zero originally necessary to calculate w΄΄ at a zero of w disappears.

Furthermore, the problem as w goes to infinity also disappears, since it turns out that the reciprocal of a solution of zww΄΄ + ww΄ − zw΄² = a − bw⁴ is a solution of zww” + ww΄ − zw΄² = b − aw⁴, so flipping back and forth between w and 1/w solves the difficulty of passing through poles when graphing the function.

If we divide both sides of the equation by w², and use the name D for the operator d/dz + zd²/dz², the equation for w becomes D ln w = aw⁻² − bw², or If we write w = p/q, it becomes D ln p − D ln q = aq²/p² − bp²/q². Since the first terms on each side have p² in their denominators, and the other two have q², this suggests considering the equations zpp΄΄ + pp΄ − zp΄² = aq² and zqq΄΄ + qq΄ − zq΄² = bp².

Since adjusting a and b can multiply the solution by any non-zero constant, let’s assume p and q have initial value 1. Then p and q are series in z whose nth degree coefficients are homogeneous nth degree polynomials in a and b, with q(a, b) = p(b, a). Calculations indicate that there are no negative coefficients and that the nth degree coefficients add up to 1/n! The latter fact is easily verified by noting that if a = b = 1, then p = q = e^z is the solution. I do not have a general proof of the former.

Finding the series inductively is easy, since the equations can be written Σ(r − s)² p_rp_s = aΣq_rq_s and Σ(r − s)² q_rq_s = bΣp_rp_s, where the left sides are summed over r + s = n with r < s, and the right sides over r + s = n − 1. We can assume an inductive hypothesis that there are no negative coefficients, but the fact that we would solve for p_n by subtracting the rest of the terms on the left from the right makes things difficult, especially since terms often cancel,. But using long integers, I have found no negatives up to degree 113, so it seems safe to make the general conjecture. Knowing that the coefficients add up to 1/n!, this would show that p and q are entire.

Calculation indicates that terms tend to congregate near the center: p(z) = 1 + az + abz²/2 + (a²b + 2ab²)z³/18 + a²b²z⁴/24 + (11a³b² + 4a²b³)z⁵/1800 + (16a⁴b² + 29a³b³)z⁶/2⁴*3⁴*5² + (29a⁴b³ + 16a³b⁴)z⁷/2⁴*3²*5²*7² + (64a⁶b² + 251a⁴b⁴)z⁸/2⁷*3⁴*5²*7² + ... In general, it appears that degree m(m − 1)/2 is the first time a term where the exponent of a differs from that of b by m/2 (if m is even) or by −(m − 1)/2 (if m is odd). The coefficient of this term is apparently 1/3²⁽ⁿ⁻²⁾*...*(2n − 3)², where n = m(m − 1)/2.

Apparent formulas for other terms increase rapidly in complexity, and I do not have a conjecture for a general formula or an idea how to prove that there are indeed no negative coefficients. So I cannot claim a proof either that p and q are entire or that w is everywhere meromorphic.

You can see graphs of w, p, and q with a = 1, b = −1 and various initial values for w. For the exact coefficients of zⁿ for n up to 113, see Pfrac.htm and for a decimal version normalized so the nth degree terms add up to 1 rather than 1/n!, see Pdec.htm.