Equation-J-1-z-Y-z-J-z-Y-1-z-2-piz-plz-Solve-this-Equation-J-z-is-First-Kind-Bessel-Function-Y-z-is-Second-Kind-Bessel-Function-aka-Neuman-Func

Question Number 194568 by MathedUp last updated on 10/Jul/23

Equation . .

J_{μ}^{(1)} (z) Y_{μ} (z) - J_{μ} (z) Y_{μ}^{(1)} (z) = - \frac{2}{π z}

plz \dots \dots Solve this Equation \dots \dots .

J_{μ} (z) is First Kind Bessel Function

Y_{μ} (z) is Second Kind Bessel Function

(aka Neuman Function)

f^{(n)} (z) n times derivate f (z) respect z

Answered by witcher3 last updated on 10/Jul/23

Y_{a} (z) = \frac{J_{a} (z) \cos (π a) - J_{- a} (z)}{\sin (π a)}

J_{a}^{'} (z) Y_{a} (z) - J_{a} (z) Y_{a}^{'} (z) =

\frac{1}{\sin (π a)} [J_{a}^{'} (z) J_{a} (z) \cos (π a) - J_{a}^{'} (z) J_{- a} (z) - J_{a} J_{a}^{'} \cos (π a)

+ J_{- a}^{^{'}} (z) J_{a} (z)]

= \frac{J_{- a}^{'} (z) J_{a} (z) - J_{a}^{'} (z) J_{- a} (z)}{\sin (π a)}

We Have J_{a} (z), J_{- a} (z) Solution of

z^{2} f^{″} (z) + {zf}^{'} (z) + (z^{2} - a^{2}) f = 0

\Rightarrow {\begin{cases} z^{2} J_{a}^{″} (z) + {zJ}_{a}^{^{'}} (z) + (z^{2} - a^{2}) J_{a} (z) = 0 \dots 1 \\ z^{2} J_{- a}^{^{″}} (z) + {zJ}_{- a}^{'} (z) + (z^{2} - a^{2}) J_{- a} (z) = 0 . . 2 \end{cases}

(1) * J_{- a} (z) - (2) * J_{a} (z)

\Leftrightarrow z^{2} (J_{a}^{″} J_{- a} - J_{- a}^{″} J_{a}) + z (J_{a}^{'} J_{- a} - J_{- a}^{^{'}} J_{a}) = 0

\Rightarrow z (J_{a}^{″} J_{- a} - J_{- a}^{″} J_{a}) + (J_{a}^{'} J_{- a} - J_{- a}^{^{'}} J_{a}) = 0

\Leftrightarrow \frac{d}{dz} (z (J_{a}^{'} (z) J_{- a} (z) - J_{- a}^{'} (z) J_{a} (z)) = 0

\Leftrightarrow J_{a}^{'} J_{- a} - J_{- a}^{'} J = \frac{c}{z}

know using taylor expension of Bassel Function

J_{a} (z) = \frac{{(\frac{z}{2})}^{a}}{Γ (1 + a)} (1 + O (z^{2}))

J_{a}^{'} (z) = \frac{{(\frac{z}{2})}^{a - 1}}{2 Γ (a)} (1 + O (z^{2}))

(J_{a}^{'} J_{- a} - J_{- a}^{^{'}} J_{a}) = \frac{1}{z} (\frac{1}{Γ (a) Γ (1 - a)} - \frac{1}{Γ (- a) Γ (1 + a)}) (1 + O (z^{2}))

Γ (z) Γ (1 - z) = \frac{π}{\sin (π z)}

we get \frac{2}{z} (\frac{\sin (π a)}{π}) \Rightarrow C = 2 \frac{\sin (π a)}{π}

\Rightarrow J_{- a}^{'} (z) J_{a} (z) - J_{a}^{'} (z) J_{- a} (z) = \frac{- 2 \sin (π a)}{π z}

J_{a}^{'} (z) Y_{a} (z) - J_{a} (z) Y_{a}^{'} (z) = \frac{1}{\sin (π a)} [J_{- a}^{'} (z) J_{a} (z) - J_{a}^{'} (z) J_{- a} (z)]

= \frac{1}{\sin (π a)} \frac{- 2 \sin (π a)}{π z} = - \frac{2}{π z}

Commented by MathedUp last updated on 11/Jul/23

WoW !!!!! I really understood after watching this Thx

Commented by MathedUp last updated on 11/Jul/23

Commented by witcher3 last updated on 11/Jul/23

happy that help you Thanx sir