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Writing y′′+P(x)y′+Q(x)y=g(x) as a Fredholm Integral Equation

  1. Jun 28, 2014 #1
    Is it possible to convert a general linear second order boundary value ode

    [tex]y'' + P(x)y' + Q(x)y = g(x), y(a) = y_a, y(b) = y_b[/tex]

    to a Fredholm integral equation, explicitly determining the Kernel in the process, without removing the [itex]y'[/itex] term? (Here is an example of doing it without the [itex]y'[/itex] term) I seem to be getting stuck.

    My answer (computed below if necessary) is

    [tex]y(x) = y_a+[\frac{y_b-y_a}{b-a}+\frac{1}{b-a}\int_a^bp(t)y(t)dt + \frac{1}{b-a}\int_a^b[q(t)-p'(t)](b-t)y(t)dt - \frac{1}{b-a}\int_a^bg(t)(b-t)dt](x-a)-\int_a^xp(t)y(t)dt - \int_a^x[q(t)-p'(t)](x-t)y(t)dt+\int_a^xg(t)(x-t)dt[/tex]

    How do we write this in terms of an integral kernel, and can we call our result a Green function?

    Note: Here is an example from Arfken of how it is done for the simple case of [itex]y''+\omega^2 y = 0, y(0)=0, y(b)=0[/itex]:


    - Computation:


    [tex]y'' = - P(x)y' - Q(x)y + g(x)[/tex]

    gives us

    [tex]y'(x) = y'_a -

    which, on getting rid of the $y'$ term by I.B.P.,

    [tex]y'(x) = y'_a +y_ap(a)-p(x)y(x) +

    gives us

    [tex]y'(x) = y'_a +y_ap(a)-p(x)y(x) -

    Integrating to find $y$ gives

    [tex]y(x) = y_a+[y'_a +y_ap(a)](x-a)-\int_a^xp(t)y(t)dt -


    [tex]y(x) = y_a+[y'_a +y_ap(a)](x-a)-\int_a^xp(t)y(t)dt -


    [tex]y(x) = y_a+[y'_a +y_ap(a)](x-a)-\int_a^xp(t)y(t)dt -

    Now, to remove the undetermined constant [itex]y'_a[/itex] we can use the B.C.
    [itex]y(b)=y_b[/itex] to find

    [tex]y'_a = \frac{y_b-y_a}{b-a}-y_ap(a)+\frac{1}{b-a}\int_a^bp(t)y(t)dt
    + \frac{1}{b-a}\int_a^b[q(t)-p'(t)](b-t)y(t)dt - \frac{1}{b-a}\int_a^bg(t)(b-t)dt[/tex]

    So that the solution is

    [tex]y(x) = y_a+[\frac{y_b-y_a}{b-a}+\frac{1}{b-a}\int_a^bp(t)y(t)dt +
    \frac{1}{b-a}\int_a^b[q(t)-p'(t)](b-t)y(t)dt -
    \frac{1}{b-a}\int_a^bg(t)(b-t)dt](x-a)-\int_a^xp(t)y(t)dt -
  2. jcsd
  3. Jul 5, 2014 #2
    maybe this will help:

    Any second order linear homogeneous ODE of the form az"+bz'+cz=0 can be written as y"-ry=0
    using the transformation [itex]y=z \exp(-\int\frac{b}{2a}dx)[/itex] with [itex]r=\frac{b^2+2ab'-2ba'-4ac}{4a^2}[/itex]

    so you transform first into something you know how to solve, then transform the answer back (in this case transforming the answer in y back to z by using the inverse transform [itex]z=y \exp(\int\frac{b}{2a}))[/itex]
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