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- 1 証明

2

Airy FunctionとModified Bessel Functionの間には以下の関係が成り立つ：

$\displaystyle \Phi(z) = \sqrt{\frac{z}{3\pi}} \, K_{1/3} \left( \frac{2}{3} z^{3/2} \right)$

(50)

1 証明

RHS of Eq.(50)	$\displaystyle = \frac{z^{1/2}}{\sqrt{3z}} \int_0^\infty e^{-\frac{2}{3} z^{3/2} \cosh(t)} \cosh\left(\frac{t}{3}\right)dt =y(z)$
$\displaystyle \di{y(z)}{z}$	$\displaystyle = \frac{z^{-1/2}}{2\sqrt{3\pi}}\frac{1}{\pi} \int_0^\infty e^{-\f... ...nfty e^{-\frac{2}{3}z^{2/3} \cosh(t)} \cosh\left(\frac{t}{3}\right) \cosh(t) dt$
	$\displaystyle = \frac{z^{-1/2}}{2\sqrt{3\pi}} K_{1/3}\left(\frac{2}{3}z^{3/2}\r... ...\right) \right) = -\frac{z}{\sqrt{3\pi}} K_{2/3}\left(\frac{2}{3}z^{3/2}\right)$	(51)
$\displaystyle \dii{y(z)}{z}$	$\displaystyle = -\frac{1}{\sqrt{3\pi}} K_{2/3}\left(\frac{2}{3}z^{3/2}\right) +... ...t) \right] = \frac{z^{3/2}}{\sqrt{3\pi}} K_{1/3}\left(\frac{2}{3}z^{3/2}\right)$
	$\displaystyle = z \sqrt{\frac{z}{3\pi}} \, K_{1/3}\left(\frac{2}{3}z^{3/2}\right) = z\Phi(z)$	(52)

これで、Eq.(50)の右辺がEq.(49)の解であることは示せた。更に、２つの境界 $z=0,z=\infty$ での両辺の値が

$\displaystyle \lim_{z\to\infty} \Phi(z)$	$\displaystyle = 0$
$\displaystyle \lim_{z\to\infty} \sqrt{\frac{z}{3\pi}} \, K_{1/3}\left(\frac{2}{3}z^{3/2}\right)$	$\displaystyle = 0$
$\displaystyle \lim_{z\to 0} \Phi(z)$	$\displaystyle = \frac{1}{\sqrt{\pi}} \int_0^\infty \cos\frac{\xi^3}{3}d\xi = \frac{3^{1/3}\sqrt{3}}{2\sqrt{\pi}} \int_0^\infty e^{t^3}dt$
	$\displaystyle = \frac{3^{1/3}}{2\sqrt{3\pi}} \int_0^\infty x^{-2/3} e^{-x} dx \, ; \quad x=t^3\to dt=\frac{1}{3}\frac{dx}{x^{2/3}}$
	$\displaystyle = \frac{3^{1/3}}{2\sqrt{3\pi}} \Gamma\left(\frac{1}{3}\right)$
$\displaystyle \lim_{z\to 0} \sqrt{\frac{z}{3\pi}} \, K_{1/3}\left(\frac{2}{3}z^{3/2}\right)$	$\displaystyle = \lim_{z\to 0} \frac{1}{\sqrt{3\pi}} z^{1/2} \frac{\Gamma\left(\... ...2}\right)^{1/3} \int_0^\infty \frac{dt}{\left(t^2+\dfrac{4}{9}z^3\right)^{5/6}}$

ここでは、

$\displaystyle K_\nu(z) = \frac{\Gamma\left(\nu+\frac{1}{2}\right) (2z)^\nu}{\sqrt{\pi}} \int_0^\infty \frac{\cos t}{\left(t^2+z^2\right)^{\nu+1/2}}dt$

の関係式を使った。更に、

	$\displaystyle \frac{1}{\sqrt{3\pi}} z^{1/2} \frac{\Gamma\left(\frac{5}{6}\right... ...ft( \frac{t}{ \left[ 4z^3/9 \right]^{1/2} } \right)}{\left(4z^3/9\right)^{5/6}}$
	$\displaystyle = \frac{\Gamma\left(\frac{5}{6}\right)}{\sqrt{3}\,\pi} \left(\fra... ...ac{1}{2}\right)\Gamma\left(\frac{1}{2}\right)}{2\Gamma\left(\frac{5}{6}\right)}$
	$\displaystyle = \frac{2^{1/3}}{\sqrt{3}\,\pi} \left(\frac{3}{2}\right)^{1/3} \f... ...frac{1}{2}\right) = \frac{3^{1/3}}{2\sqrt{3\pi}} \Gamma\left(\frac{1}{3}\right)$

となる。よって２つの境界での両辺の値が等しいことが分かる。二階の微分方程式の解の一意性から、２つの境界条件を与えれば、解が一意に決まることにより、 Eq.(50)が正しいことが証明された。

著者: 茅根裕司 chinone_at_astr.tohoku.ac.jp