Chandrasekhar's H-function

Explicit solution in the complex plane

In complex variable ${\displaystyle z}$ the H equation is

${\displaystyle H(z)=1-{\int }_0^1\frac{z}{z+\mu }H(\mu )\mathrm{\Psi }(\mu )d\mu ,{\int }_0^1|\mathrm{\Psi }(\mu )|d\mu \le \frac{1}{2},{\int }_0^{\delta }|\mathrm{\Psi }(\mu )|d\mu \to 0,\delta \to 0}$

then for ${\displaystyle \mathrm{\Re }(z)>0}$, a unique solution is given by

${\displaystyle \ln H(z)=\frac{1}{2\pi i}{\int }_{-i\mathrm{\infty }}^{+i\mathrm{\infty }}\ln T(w)\frac{z}{w^2-z^2}dw}$

where the imaginary part of the function ${\displaystyle T(z)}$ can vanish if ${\displaystyle z^2}$ is real i.e., ${\displaystyle z^2=u+iv=u(v=0)}$. Then we have

${\displaystyle T(z)=1-2{\int }_0^1\mathrm{\Psi }(\mu )d\mu -2{\int }_0^1\frac{{\mu }^2\mathrm{\Psi }(\mu )}{u-{\mu }^2}d\mu }$

The above solution is unique and bounded in the interval ${\displaystyle 0\le z\le 1}$ for conservative cases. In non-conservative cases, if the equation ${\displaystyle T(z)=0}$ admits the roots ${\displaystyle \pm 1/k}$, then there is a further solution given by

${\displaystyle H_1(z)=H(z)\frac{1+kz}{1-kz}}$

Properties

• ${\displaystyle {\int }_0^{1}H(\mu )\mathrm{\Psi }(\mu )d\mu =1-{\left[1-2{\int }_0^1\mathrm{\Psi }(\mu )d\mu \right]}^{1/2}}$. For conservative case, this reduces to ${\displaystyle {\int }_0^1\mathrm{\Psi }(\mu )d\mu =\frac{1}{2}}$.
• ${\displaystyle {\left[1-2{\int }_0^1\mathrm{\Psi }(\mu )d\mu \right]}^{1/2}{\int }_0^{1}H(\mu )\mathrm{\Psi }(\mu ){\mu }^{2}d\mu +\frac{1}{2}{\left[{\int }_0^{1}H(\mu )\mathrm{\Psi }(\mu )\mu d\mu \right]}^2={\int }_0^1\mathrm{\Psi }(\mu ){\mu }^{2}d\mu }$. For conservative case, this reduces to ${\displaystyle {\int }_0^{1}H(\mu )\mathrm{\Psi }(\mu )\mu d\mu ={\left[2{\int }_0^1\mathrm{\Psi }(\mu ){\mu }^{2}d\mu \right]}^{1/2}}$.
• If the characteristic function is ${\displaystyle \mathrm{\Psi }(\mu )=a+b{\mu }^2}$, where ${\displaystyle a,b}$ are two constants(have to satisfy ${\displaystyle a+b/3\le 1/2}$) and if ${\displaystyle {\alpha }_n={\int }_0^{1}H(\mu ){\mu }^{n}d\mu ,n\ge 1}$ is the nth moment of the H function, then we have

${\displaystyle {\alpha }_0=1+\frac{1}{2}(a{\alpha }_0^2+b{\alpha }_1^2)}$

and

${\displaystyle (a+b{\mu }^2){\int }_0^1\frac{H({\mu }^{\prime })}{\mu +{\mu }^{\prime }}d{\mu }^{\prime }=\frac{H(\mu )-1}{\mu H(\mu )}-b({\alpha }_1-\mu {\alpha }_0)}$