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Hill Determinant

A Determinant which arises in the solution of the second-order Ordinary Differential Equation

x^2{d^2\psi\over dx^2}+x{d\psi\over dx}+\left({{\textstyle{1...
...x^2+{\textstyle{1\over 2}}h^2-b+{h^2\over 4x^2}}\right)\psi=0.
\end{displaymath} (1)

Writing the solution as a Power Series
\psi=\sum_{n=-\infty}^\infty a_n x^{s+2n}
\end{displaymath} (2)

gives a Recurrence Relation
h^2a_{n+1}+[2h^2-4b+16(n+{\textstyle{1\over 2}}s)^2]a_n+h^2a_{n-1}=0.
\end{displaymath} (3)

The value of $s$ can be computed using the Hill determinant
\ddots & \vdots & \vdots & \vdo...
...u}& \vdots & \vdots & \vdots & \vdots & \ddots\cr}\right\vert,
\end{displaymath} (4)

$\displaystyle \sigma$ $\textstyle =$ $\displaystyle {\textstyle{1\over 2}}s$ (5)
$\displaystyle \alpha^2$ $\textstyle =$ $\displaystyle {\textstyle{1\over 4}}b-{\textstyle{1\over 8}} h^2$ (6)
$\displaystyle \beta$ $\textstyle =$ $\displaystyle {\textstyle{1\over 4}}h,$ (7)

and $\sigma$ is the variable to solve for. The determinant can be given explicitly by the amazing formula
\Delta(s)=\Delta(0)-{\sin^2(\pi s/2)\over\sin^2({\textstyle{1\over 2}}\pi\sqrt{b-{\textstyle{1\over 2}}h^2}\,)},
\end{displaymath} (8)


\Delta(0)=\left\vert\matrix{\ddots & \vdots & \vdots & \vdot...
...\vdots & \vdots & \vdots & \vdots & \ddots\cr}\right\vert,eqno
\end{displaymath} (9)

leading to the implicit equation for $s$,

\sin^2({\textstyle{1\over 2}}\pi s)=\Delta(0)\sin^2\left({{\...
...yle{1\over 2}}\pi\sqrt{b-{\textstyle{1\over 2}}h^2}\,}\right).
\end{displaymath} (10)

See also Hill's Differential Equation


Morse, P. M. and Feshbach, H. Methods of Theoretical Physics, Part I. New York: McGraw-Hill, pp. 555-562, 1953.

© 1996-9 Eric W. Weisstein