Klasa funkcji cylindrycznych

Funkcje Hankela pierwszego rodzaju:

$${H_{\nu}}^{(1)}(x)=\frac{i}{\sin \pi \nu}\big[ J_{\nu}(x) e^{-i \pi \nu}- J_{-\nu}(x)\big], \quad \nu \neq n,$$

$${H_{n}}^{(1)}(x)=J_{n}(x)+\frac{i}{\pi}\bigg[\frac{\partial J_{\... ... \nu} +{(-1)}^{n+1} \frac{\partial J_{-\nu}(x)}{\partial \nu}\bigg]_{\nu=n};$$

Funkcje Hankela drugiego rodzaju:

$${H_{\nu}}^{(2)}(x)=\frac{1}{i \sin \pi \nu}\big[ J_{\nu}(x) e^{i \pi \nu}- J_{-\nu}(x)\big], \quad \nu \neq n,$$

$${H_{n}}^{(2)}(x)=J_{n}(x)-\frac{i}{\pi}\bigg[\frac{\partial J_{\... ... \nu} +{(-1)}^{n+1} \frac{\partial J_{-\nu}(x)}{\partial \nu}\bigg]_{\nu=n};$$

Funkcje Neumanna

$$N_{\nu}(x)=\frac{1}{\sin \pi \nu}\big[ J_{\nu}(x) \cos \pi \nu- J_{-\nu}(x)\big], \quad \nu \neq n,$$

$$N_{n}(x)=\frac{1}{\pi}\bigg[\frac{\partial J_{\nu}(x)}{\partial \nu} +{(-1)}^{n+1} \frac{\partial J_{-\nu}(x)}{\partial \nu}\bigg]_{\nu=n};$$

Funkcje argumentu urojonego

$$I_{\nu}(x)= e^{-\frac{\pi \nu i}{2}}J_{\nu}(ix), \quad K_{\nu}(x)=\frac{\pi i}{2} e^{\frac{\pi \nu i}{2}}{H_{\nu}}^{(1)}(ix),$$

tj.

$${H_{\nu}}^{(1)}(x)=J_{\nu}(x) + i N_{\nu}(x),$$

$${H_{\nu}}^{(2)}(x)=J_{\nu}(x) - i N_{\nu}(x),$$

Przy $x \to \infty$ zachodzą wzory asymptotyczne:

$${H_{\nu}}^{(1)}(x)\simeq \sqrt{\frac{2}{\pi x}} e^{i(x -\frac{\pi}{2}\nu - \frac{\pi}{4})}+O\big({x}^{-\frac{3}{2}}\big),$$

$${H_{\nu}}^{(2)}(x)\simeq \sqrt{\frac{2}{\pi x}} e^{-i(x -\frac{\pi}{2}\nu - \frac{\pi}{4})}+O\big({x}^{-\frac{3}{2}}\big),$$

$$N_{\nu}(x)\simeq \sqrt{\frac{2}{\pi x}} \sin{\big(x -\frac{\pi}{2}\nu - \frac{\pi}{4}\big)}+O\big({x}^{-\frac{3}{2}}\big),$$

$$I_{\nu}(x) \simeq \frac{e^{x}}{\sqrt{2 \pi x}} \Big[1+O\big({x}^{-1} \big) \Big],$$

$$K_{\nu}(x) \simeq \sqrt{\frac{\pi}{2 x}} e^{-x} \Big[1+O\big({x}^{-1} \big) \Big],$$

$${H_{0}}^{(1)}(x)\simeq -\frac{2i}{\pi}\ln\frac{1}{x} + \ldots, \quad {H_{0}}^{(2)}(x)\simeq \frac{2i}{\pi}\ln\frac{1}{x} + \ldots,$$

$$N_{0}(x)\simeq -\frac{2}{\pi}\ln\frac{1}{x} + \ldots,\quad K_{0}(x)\simeq \ln\frac{1}{x} + \ldots,$$

Ćwiczenie 13.2   Wykazać, że w sensie uogólnionym

$$\textrm{a) }e^{\frac{x}{2}(t-\frac{1}{t})}= \sum_{n \in \mathbb{Z}}J_{n}(x)t^{n};$$ (13.36)

$$\textrm{b) } J_{n}(x)=\frac{1}{2 \pi} \int_{-\pi}^{\pi}e^{i(n \nu - x \sin \nu)} \mathrm{d}\nu$$ (13.37)