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14.26.7 The scalar coefficients68DF63E92045

On the central interval $-1\leq r\leq 1$, the coefficients will make a different rank-one factor vanish in each of the two solution forms. Outside a larger interval they coincide with the Fubini–Study coefficients, so the resulting forms extend across the omitted projective lines. Primes denote derivatives with respect to $r$.

Choose an even smooth function $\beta $ supported in $[-1,1]$, strictly positive on $(-1,1)$, with $\beta '>0$ on $(-1,0)$, and normalized by $\int _{-1}^1\beta =2$; for instance, a positive multiple of $\exp (-1/(1-r^2))$ on $(-1,1)$, extended by zero. Define

\[ F(r)=-1+\int _{-1}^ r\beta (t)\, \mathrm{d}t\quad (-1<r<1), \qquad F=-1\ (r\leq -1),\quad F=1\ (r\geq 1), \]

and put $p=\beta /2$, $q=\beta '/2$. Then $F$ is smooth and odd, $p$ is even, and $q$ is odd, positive on $(-1,0)$ and negative on $(0,1)$. Both $p$ and $q$ vanish to infinite order at the endpoints.

Take smooth $a\geq 0$ with $a=1$ on $(-\infty ,-2/3]$ and $a=0$ on $[-1/3,\infty )$, and put $b(r)=a(-r)$. Choose an even nonnegative $\chi \in C_ c^\infty ((-1/3,1/3))$ with $\chi (0)>0$. On $[-1,1]$ set

\[ z_0=(q^2+\chi ^2)^{1/2}. \]

It is smooth: the expression under the square root is positive in the interior, and near each endpoint $z_0=|q|$, where $q$ has a fixed sign. It is positive in the interior and vanishes to infinite order at the endpoints. Also,

6C8664C40E89 14.26.8
\begin{equation} z_0\geq |q|,\qquad a(z_0-q)=0,\qquad b(z_0+q)=0. \end{equation}

The first cancellation holds because $\chi =0$ on the support of $a$ and $q\geq 0$ there; the second follows by reflection.

On $(-\infty ,-1]$ choose smooth functions $\ell ,s$ such that $0\leq \ell \leq 1$, $\ell =0$ on $(-\infty ,-2]$, $\ell =1$ on $[-3/2,-1]$, while $s>0$ for $r<-1$, $s=1$ on $(-\infty ,-3/2]$, and $s$ vanishes to infinite order at $-1$. Such functions are easily obtained from smooth cutoff functions and a function like $\exp (-1/(r+1)^2)$. Define

\[ \begin{aligned} x& =(1-\ell )\sigma +\ell ,& y& =(1-\ell )\sigma (-r)+\ell s,& z& =(1-\ell )\sigma '+\ell s & & (r\leq -1),\\ x& =p+a,& y& =p+b,& z& =z_0 & & (-1\leq r\leq 1), \end{aligned} \]

and for $r\geq 1$ use $x(r)=y(-r)$, $y(r)=x(-r)$, $z(r)=z(-r)$.

The definitions join smoothly: at $-1$ both definitions give the values $(1,0,0)$, and their differences vanish to infinite order there, since $a=1$ and $b=0$ near $-1$. The central definitions are symmetric under reflection, so the same holds at $1$. We have

\[ x,y,z>0\quad \hbox{for }r\ne \pm 1,\qquad (x,y,z)(-1)=(1,0,0),\quad (x,y,z)(1)=(0,1,0). \]

For $|r|\geq 2$ their values are $(\sigma (r),\sigma (-r),\sigma '(r))$.

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