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I BVP vs IVP when solving for eigenvalues

  1. Dec 7, 2016 #1
    I am being asked to find ##\lambda## such that ##y'' + \lambda y = 0; ~y(0) =0; ~y'( \pi ) = 0##. This is an eigenvalue problem where we are given boundary conditions. ##\lambda## can be found such that we don't have a trivial solution if we test the different cases when ##\lambda < 0, \lambda = 0##, or ##\lambda > 0##, and solve for what ##\lambda## must be such that we have non-trivial solutions.

    I'm just curious as to why these problems are given with boundary conditions instead of initial conditions. Is there some theory behind why we need boundary conditions in order to determine the eigenvalues? Would being given initial conditions allow us to do the same thing?
     
  2. jcsd
  3. Dec 7, 2016 #2
    Looks to me like a second-order linear differential equation. It also looks to me like an initial value problem, not a boundary value problem. Boundary value would be if you had ##y (x_1) = y_1,~ y(x_2) = y_2##. That is just one of the methods of solving for it.
     
  4. Dec 8, 2016 #3

    joshmccraney

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    Are you sure, TJGlib? Boundary value problems are specified at the boundaries; the 2nd order ODE as specified by Mr Davis 97 is specified ##0## and at ##\pi## and is then by definition a BVP, hence NOT an IVP. The derivatives of the boundary conditions do not dictate whether a problem is a BVP or IVP; but rather where the boundary conditions are specified.
     
  5. Dec 8, 2016 #4

    lurflurf

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    boundary value problems (bvp) tend to be more versatile than initial value problems (ivp)
    for example often for a situation an ivp has a unique solution
    while a bvp may have a unique solution, infinite solutions, or no solution

    often the bvp has arisen naturally
    sometimes we convert the bvp to an ivp by finding one with the same solution
    ie shooting method

    often it is not possible or desirable to do this
    in fact it is often helpful to convert an ivp to a bvp
    for example if the ivp is sensitive a bvp may be less so

    many methods involve breaking a function or region into pieces solving the parts and reassembling them
    this can be done for both ivp and bvp but often bvp are more convenient
    scale is often a problem in ivp such is in boundary layers and other situations when different effects dominate different regions
     
  6. Dec 8, 2016 #5

    It's instructive to see what happen when we to solving the similar initial value problem: Find [itex] \lambda [/itex] such that [itex] y'' +\lambda y =0 [/itex] for [itex] y(0)=0 [/itex] and [itex] y'(0)=0 [/itex].

    For [itex] \lambda = 0 [/itex] the general solution is [itex] y= ax +b [/itex] but when we apply the boundary conditions we find that [itex] a=0, b=0 [/itex].

    For [itex] \lambda > 0 [/itex] the general solution is [itex] y= a sin \sqrt{\lambda} x +b cos \sqrt{\lambda} x[/itex], but when we apply the boundary conditions we again find that [itex] a=0, b=0 [/itex].

    For [itex] \lambda < 0 [/itex] the general solution is [itex] y= a exp (\sqrt{-\lambda} x) +b exp (-\sqrt{-\lambda} x)[/itex], but when we apply the boundary conditions we again find that [itex] a=0, b=0 [/itex].

    You can see that in this case the initial value eigenvalue problem has zero nontrivial solutions.
     
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