Equation.. J_𝛍 ^((1)) (z)Y_𝛍 (z)−J_𝛍 (z)Y_𝛍 ^((1)) (z)=−(2/(πz)) plz......Solve this Equation....... 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


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

$Equation . .$ $J_{μ}^{(1)} (z) Y_{μ} (z) - J_{μ} (z) Y_{μ}^{(1)} (z) = - \frac{2}{π z}$ $plz . . . . . . Solve this Equation . . . . . . .$ $J_{μ} (z) is First Kind Bessel Function$ $Y_{μ} (z) is Second Kind Bessel Function$ $(aka Neuman Function)$ $f^{(n)} (z) n times derivate f (z) respect z$
	Answered by witcher3 last updated on 10/Jul/23
	$Y_a (z)=((J_a (z)cos(πa)−J_(−a) (z))/(sin(πa))) J_a ′(z)Y_a (z)−J_a (z)Y′_a (z)= (1/(sin(πa)))[J′_a (z)J_a (z)cos(πa)−J′_a (z)J_(−a) (z)−J_a J′_a cos(πa) +J_(−a) ^′ (z)J_a (z)] =((J′_(−a) (z)J_a (z)−J′_a (z)J_(−a) (z))/(sin(πa))) We Have J_a (z),J_(−a) (z) Solution of z^2 f′′(z)+zf′(z)+(z^2 −a^2 )f=0 ⇒ { ((z^2 J_a ′′(z)+zJ_a ^′ (z)+(z^2 −a^2 )J_a (z)=0...1)),((z^2 J_(−a) ^(′′) (z)+zJ′_(−a) (z)+(z^2 −a^2 )J_(−a) (z)=0..2)) :} (1)∗J_(−a) (z)−(2)∗J_a (z) ⇔z^2 (J′′_a J_(−a) −J′′_(−a) J_a )+z(J′_a J_(−a) −J_(−a) ^′ J_a )=0 ⇒z(J′′_a J_(−a) −J′′_(−a) J_a )+(J′_a J_(−a) −J_(−a) ^′ J_a )=0 ⇔(d/dz)(z(J_a ′(z)J_(−a) (z)−J′_(−a) (z)J_a (z))=0 ⇔J′_a J_(−a) −J′_(−a) J=(c/z) know using taylor expension of Bassel Function J_a (z)=((((z/2))^a )/(Γ(1+a)))(1+O(z^2 )) J′_a (z)=((((z/2))^(a−1) )/(2Γ(a)))(1+O(z^2 )) (J′_a J_(−a) −J_(−a) ^′ J_a )=(1/z)((1/(Γ(a)Γ(1−a)))−(1/(Γ(−a)Γ(1+a))))(1+O(z^2 )) Γ(z)Γ(1−z)=(π/(sin(πz))) we get (2/z)(((sin(πa))/π))⇒C=2((sin(πa))/π) ⇒J′_(−a) (z)J_a (z)−J′_a (z)J_(−a) (z)=((−2sin(πa))/(πz)) J_a ′(z)Y_a (z)−J_a (z)Y′_a (z)=(1/(sin(πa)))[J′_(−a) (z)J_a (z)−J′_a (z)J_(−a) (z)] =(1/(sin(πa))) ((−2sin(πa))/(πz))=−(2/(πz))$
	$Y_{a} (z) = \frac{J_{a} (z) \cos (π a) - J_{- a} (z)}{\sin (π a)}$ $J_{a}^{'} (z) Y_{a} (z) - J_{a} (z) Y_{a}^{'} (z) =$ $\frac{1}{\sin (π a)} [J_{a}^{'} (z) J_{a} (z) \cos (π a) - J_{a}^{'} (z) J_{- a} (z) - J_{a} J_{a}^{'} \cos (π a)$ $+ J_{- a}^{^{'}} (z) J_{a} (z)]$ $= \frac{J_{- a}^{'} (z) J_{a} (z) - J_{a}^{'} (z) J_{- a} (z)}{\sin (π a)}$ $We Have J_{a} (z), J_{- a} (z) Solution of$ $z^{2} f^{″} (z) + {zf}^{'} (z) + (z^{2} - a^{2}) f = 0$ $\Rightarrow {\begin{cases} z^{2} J_{a}^{″} (z) + {zJ}_{a}^{^{'}} (z) + (z^{2} - a^{2}) J_{a} (z) = 0 . . . 1 \\ z^{2} J_{- a}^{^{″}} (z) + {zJ}_{- a}^{'} (z) + (z^{2} - a^{2}) J_{- a} (z) = 0 . . 2 \end{cases}$ $(1) * J_{- a} (z) - (2) * J_{a} (z)$ $\Leftrightarrow z^{2} (J_{a}^{″} J_{- a} - J_{- a}^{″} J_{a}) + z (J_{a}^{'} J_{- a} - J_{- a}^{^{'}} J_{a}) = 0$ $\Rightarrow z (J_{a}^{″} J_{- a} - J_{- a}^{″} J_{a}) + (J_{a}^{'} J_{- a} - J_{- a}^{^{'}} J_{a}) = 0$ $\Leftrightarrow \frac{d}{dz} (z (J_{a}^{'} (z) J_{- a} (z) - J_{- a}^{'} (z) J_{a} (z)) = 0$ $\Leftrightarrow J_{a}^{'} J_{- a} - J_{- a}^{'} J = \frac{c}{z}$ $know using taylor expension of Bassel Function$ $J_{a} (z) = \frac{{(\frac{z}{2})}^{a}}{Γ (1 + a)} (1 + O (z^{2}))$ $J_{a}^{'} (z) = \frac{{(\frac{z}{2})}^{a - 1}}{2 Γ (a)} (1 + O (z^{2}))$ $(J_{a}^{'} J_{- a} - J_{- a}^{^{'}} J_{a}) = \frac{1}{z} (\frac{1}{Γ (a) Γ (1 - a)} - \frac{1}{Γ (- a) Γ (1 + a)}) (1 + O (z^{2}))$ $Γ (z) Γ (1 - z) = \frac{π}{\sin (π z)}$ $we get \frac{2}{z} (\frac{\sin (π a)}{π}) \Rightarrow C = 2 \frac{\sin (π a)}{π}$ $\Rightarrow J_{- a}^{'} (z) J_{a} (z) - J_{a}^{'} (z) J_{- a} (z) = \frac{- 2 \sin (π a)}{π z}$ $J_{a}^{'} (z) Y_{a} (z) - J_{a} (z) Y_{a}^{'} (z) = \frac{1}{\sin (π a)} [J_{- a}^{'} (z) J_{a} (z) - J_{a}^{'} (z) J_{- a} (z)]$ $= \frac{1}{\sin (π a)} \frac{- 2 \sin (π a)}{π z} = - \frac{2}{π z}$
	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

	$happy that help you Thanx sir$
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