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Orthogonality Property of Hyperbolic functions ?

  1. Apr 28, 2009 #1
    Orthogonality Property of Hyperbolic functions ???

    Hi all,
    I have seen Orthogonal property for trigonomeric functions but I am unsure if there is something similar for sinh() , cosh() ? . I know that the integral of inner product of the two functions should be zero for them to be called as orthogonal.

    I am asking this question because if one applies Variable separable form of F(x)G(y) to solve a laplace equation, then one always gets one of F(x) or G(y) to be hyperbolic. With this being the case ,doesn any body know how to extract coefficients from series form of solution
    of laplace equation when (say) only sinh(kx) is remaining :


    for e.g Summation {Pn cosh(kx) } =phi . How to extract Pn ?


    Had it been Summation {Pn cos(kx) } =phi ,then one can use orthogonal property of trigonometric function .

    I hope ,I am clear.

    Thanks ,
    A.S
     
  2. jcsd
  3. May 13, 2009 #2
    Re: Orthogonality Property of Hyperbolic functions ???

    I think most of the orthogonal functions are obtained from the solutions of the Sturm–Liouville eigenvalue equations

    [tex](py')'+(q+\lambda r)y=0[/tex]

    where [tex]\lambda[/tex] is a constant.

    The trigonometric and hyperbolic come from solving the following equation
    [tex]y''+\lambda y=0[/tex].
    As a general solution, we obtain trigometric solution if [tex]\lambda > 0 [/tex] and hyperbolic solution if [tex]\lambda < 0 [/tex]. But if we further impose the boundary conditions then for [tex]\lambda < 0 [/tex] the solution is trivial y=0.

    So I suspect that there is no such thing as orthogonal property for hyperbolic functions.
     
  4. May 13, 2009 #3

    djeitnstine

    User Avatar
    Gold Member

    Re: Orthogonality Property of Hyperbolic functions ???

    This is true.
     
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