Let \(\phi\) be holomorphic, and let \(u_1,u_2\) be linearly independent solutions of \[ u''+\phi u=0. \]
Show that the Wronskian \[ W(z)=\det \begin{pmatrix} u_1&u_2\\ u_1'&u_2' \end{pmatrix} \] is constant.
Let \(f=u_1/u_2\). Show that \[ S_f=C\phi \] for a universal constant \(C\), determine \(C\), and explain why \(S_f\) is independent of the chosen basis \(u_1,u_2\) of the solution space.
Solution
The Wronskian is constant, and \[ \boxed{S_f=2\phi}. \] Thus \(C=2\).
1. The Wronskian\[ W=u_1u_2'-u_2u_1' \]is constant.
Differentiate: \[ \begin{aligned} W' &=u_1'u_2'+u_1u_2''-u_2'u_1'-u_2u_1''\\ &=u_1u_2''-u_2u_1''. \end{aligned} \] Since both solutions satisfy \[ u_j''=-\phi u_j, \] we get \[ W' =-\phi u_1u_2+\phi u_2u_1 =0. \] Hence \(W\) is constant. Because \(u_1,u_2\) are linearly independent, this constant is nonzero.
2. On every open set where \(u_2\ne0\), the quotient \(f=u_1/u_2\) satisfies\[ N_f=-2\frac{u_2'}{u_2}. \]
Using the sign convention for \(W\) in 1, \[ f' =\frac{u_1'u_2-u_1u_2'}{u_2^2} =-\frac{W}{u_2^2}. \] Since \(W\) is a nonzero constant, \[ N_f =\frac{f''}{f'} =\frac{d}{dz}\log f' =-2\frac{u_2'}{u_2}. \]
3. The Schwarzian is\[ S_f=2\phi. \]
From 2, \[ N_f' =-2\left(\frac{u_2''}{u_2}-\left(\frac{u_2'}{u_2}\right)^2\right), \] while \[ \frac12N_f^2 =2\left(\frac{u_2'}{u_2}\right)^2. \] Therefore \[ \begin{aligned} S_f &=N_f'-\frac12N_f^2\\ &=-2\frac{u_2''}{u_2}\\ &=2\phi, \end{aligned} \] because \(u_2''=-\phi u_2\). This identity holds wherever \(u_2\ne0\), and hence throughout by analytic continuation of the resulting quadratic differential.
4. The Schwarzian does not depend on the chosen basis of solutions.
If \((v_1,v_2)\) is another basis, then for constants \(a,b,c,d\) with \(ad-bc\ne0\), \[ \frac{v_1}{v_2} =\frac{a u_1+b u_2}{c u_1+d u_2} =M\!\left(\frac{u_1}{u_2}\right), \] where \[ M(w)=\frac{aw+b}{cw+d} \] is a Möbius transformation. Möbius transformations have zero Schwarzian. Hence the Schwarzian chain rule gives \[ S_{M\circ f} =S_f+(S_M\circ f)(f')^2 =S_f. \] Equivalently, 3 already shows that every basis gives the same intrinsic value \(2\phi\).