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Solving an eigenvalue equation with boundary conditions

  1. May 4, 2015 #1
    Suppose that we want to solve the eigenvalue equation with Dirichlet boundary conditions

    ## \bigg(-\frac{d^2}{dx^2}+V(x)\bigg) \phi_n = \lambda_n \phi_n,\ \ \ \ \ \ \ \ \ \ \ \ \ \phi_n(0)=0,\ \phi_n(1)=0, ##

    where ##0 < \lambda_1 < \lambda_2 < ...## are discrete, non-degenerate eigenvalues that are bounded from below.

    I've read somewhere that we can achieve this by constructing an initial value problem with the same operator, but with different boundary conditions:

    ##\bigg(-\frac{d^2}{dx^2}+V(x)\bigg) u_\lambda = \lambda u_\lambda,\ \ \ \ \ \ \ \ \ \ \ \ \ u_\lambda(0)=0,\ u^{'}_\lambda(1)=0.##

    I can see that this determines the function ##u_\lambda (x)## uniquely, because the second order differential equation is supplanted with two boundary conditions. Also, I understand that if ##\lambda## is to be an eigenvalue of the original problem, then it also holds that ##u_\lambda(1)=0##. That much I understand.

    However, I don't understand why we need to convert the boundary value problem into an initial value problem in order to solve it, and how the condition ##u^{'}_\lambda(1)=0## comes about anyway.

    Can somebody please explain? :(
     
  2. jcsd
  3. May 6, 2015 #2

    Simon Bridge

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    ... you don't need[\i] to. The trouble is that the V(x) can be arbitrarily complicated so it is a good idea to have a bunch of different approaches available for use.
    ... perhaps it will help you to relate the ##u##'s to the ##\phi##'s.
     
  4. May 6, 2015 #3
    There are a few false statements in your post. Do you remember where you read about this method.

    With the boundary conditions that you list the problem is still an boundary value problem not an initial value problem. Even worse it will give you the wrong eigenvalues.

    In order for the problem to be an initial value problem the boundary conditions need to be something like: [itex] u(0)=0[/itex] and [itex] u'(0) =1 [/itex], where both boundary conditions apply at the same "initial" point.

    Actually it does not. If you multiply u by a constant then you will get a new function that is also a solution.


    Perhaps you read about shooting methods. Shooting methods are numerical techniques frequently used to find solutions to BVP by converting them to initial value problems. To do this you have to guess one of the initial conditions and you ignore the boundary condition and "end boundary. Then you iterate on that guess by solving the initial value problem multiple times until you find the right guess to satisfy the "end" boundary condition.

    You can tweak this for an eigenvalue problem by asserting that the guess initial value has a value of 1. You then iterate on the eigenvalues to find solutions of the original boundary value problem. These works because the eigenfunctions are not unique. Therefore you can rescale the equation such that the "guess" initial value has a value of 1.
     
  5. May 9, 2015 #4
    I've read it on page 22 of the pdf file named 'Functional determinants in quantum field theory' on this page http://www.itp.uni-hannover.de/saalburg/2008.php.

    Actually, the boundary conditions are indeed ##u_\lambda(0)=0,\ u^{'}_\lambda(0)=1.## That was my mistake.

    However, I am having a bit of trouble understanding the reason for the boundary condition ##u^{'}_\lambda(0)=1.## The footnote mentions that the choice of the second condition in is just a choice of normalization for u, and that a different choice of normalization would render a multiplicative factor.

    I don't quite understand what that means. :(
     
  6. May 9, 2015 #5

    pasmith

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    The point is that your eigenfunction [itex]u[/itex] must be a solution which is not constantly zero. You can ensure that by taking [itex]u'(0) \neq 0[/itex].

    Now if [itex]v[/itex] is a solution of [tex]
    -\frac{d^2v}{dx^2} + V(x)v = \lambda v\qquad(*)
    [/tex] which satisfies the initial conditions [itex]v(0) = 0[/itex] and [itex]v'(0) = k[/itex] for non-zero [itex]k[/itex], then [itex]w(x) = v(x)/k[/itex] is a solution of (*) which satisfies the initial conditions [itex]w(0) = 0[/itex] and [itex]w'(0) = 1[/itex].

    Thus there is nothing lost by taking [itex]u'(0) = 1[/itex].

    EDIT: In a normal shooting method, [itex]\lambda[/itex] would be known and fixed and you would be adjusting [itex]u'(0)[/itex] until [itex]u(1) = 0[/itex]. Here [itex]u'(0) = 1[/itex] is fixed and you are adjusting [itex]\lambda[/itex] until [itex]u(1) = 0[/itex].
     
    Last edited: May 9, 2015
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