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Homework Help: A Bessel's functions of the second kind (Neumann' functions) deduction

  1. Mar 18, 2013 #1
    1. The problem statement, all variables and given/known data
    I need to obtain the Bessel functions of the second kind, from the expressions of the Bessel functions of the first kind.

    2. Relevant equations
    Laplace equation in circular cylindrical coordinates reads
    [tex]\nabla^2\phi(\rho,\varphi,z)=0[/tex] with [tex]\nabla^2=\frac{1}{\rho}\frac{\partial}{\partial \rho}\left(\rho \frac{\partial}{\partial \rho} \right)+\frac{1}{\rho^2}\frac{\partial^2}{\partial \varphi^2}+\frac{\partial^2}{\partial z^2}[/tex]

    3. The attempt at a solution
    Supposing that [tex]\phi(\rho,\varphi,z)=R(\rho)\Phi(\varphi)Z(z)[/tex] i get [tex]\begin{eqnarray} Z_{m}(z)=A_{1_m}e^{m z}+A_{2_m}e^{-m z} && && \Phi_{k}(\varphi)=A_{3_k}\cos{k \varphi}+A_{4_k}\sin{k\varphi} \end{eqnarray}[/tex] with [itex]m[/itex] and [itex]k[/itex] being the eigenvalues from the ODE's associated with Z and [itex]\Phi[/itex], respectively.
    Then the associated ODE for the radial component is [tex]\frac{d^2R}{d(k\rho)^2}+\frac{1}{k\rho}\frac{dR}{d(k\rho)}+\left(1-\frac{m^2}{(k\rho)^2}\right)R=0[/tex] and defining [itex]x=k\rho[/itex] i get [tex]\frac{d^2R}{d x^2}+\frac{1}{x}\frac{dR}{d x}+\left(1-\frac{m^2}{x^2}\right)R=0[/tex] which is Bessel's differential equation. Using Frobenius method to remove the singularity, I'm able to compute Bessel functions of the first kind [tex]J_{m}(x)=\sum_{s=0}^{\infty}{\frac{(-1)^s}{s!(m+s)!}\left(\frac{x}{2}\right)^{m+2s}}[/tex] for m integer. Changing n by -n and removing the zero terms in the series for [itex]J_{-m}(x)[/itex], we see that
    [tex]J_{-m}(x)=(-1)^{m}J_{m}(x)[/tex] so both functions are linearly dependent, and we need to specify another function of x linearly independent of [itex]J_{m}(x)[/itex] because the differential equation is a second order one.
    Then, a common way to find a second function of a differential equation of the form [tex]y''(x)+p(x)y'(x)+q(x)y(x)=0[/tex] with a known solution [itex]y_{1}(x)[/itex], we can suppose that [itex]y_{2}(x)=y_{1}(x)g(x)[/itex]. After deriving [itex]y_{2}[/itex] and substituting in the differential equation, we get
    but if i try to use this method to find the Neumann's functions [itex]Y_{m}(x)[/itex], i cannot find the expression
    [tex]Y_{m}(x)=\frac{\cos{mx}J_{m}(x)-J_{-m}(x)}{\sin{mx}}[/tex] which commonly one finds that it is a "definition" for the Neumann's functions. Any ideas?
  2. jcsd
  3. Mar 19, 2013 #2

    I'm sorry, but the last expression was [tex]Y_{m}(x)=\frac{\cos{m\pi}J_{m}(x)-J_{-m}(x)}{\sin{m\pi}}[/tex] Any kind of help would be greatly appreciated!
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