Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Volterra Eqn of 2nd Kind -> DEQ

  1. Mar 18, 2007 #1
    Volterra Eqn of 2nd Kind --> DEQ

    1. The problem statement, all variables and given/known data
    I need to convert y(x) = 1 - x + int[dt(x-t)y(t)] from 0 to x to a differential equation with the appropriate boundary conditions.

    3. The attempt at a solution

    OK I just had a problem converting a DEQ into an integral equation so I know the form it will take, I know it will be a 2nd order homogeneous equation... something like d^2y/dx^2 + y(x) = 0 with y(a) = b , dy/dx|c = d for some constants. I know it won't be a periodic boundary condition because that turns into a Fredholm Equation of the 2nd Kind.

    So I applied d/dx to both sides getting

    dy/dx = -1 + int[dty(t)] from 0 to x

    I think this is valid since the integrand is with respect to t, I just applied the differentiation within the integral.

    I'm not sure what to do from here though.
  2. jcsd
  3. Mar 18, 2007 #2
    Ok so I missed the obvious step of solving for the boundary conditions by simply plugging in x= 0 into my y(x) and dy/dx equations

    so y(0) = 1 and dy(0)/dx = -1

    This is what I got from my sketchy inspection method prior to this. I know the answer now, but I still don't know how to show it will become a 2nd order homogeneous equation.
  4. Mar 18, 2007 #3


    User Avatar
    Science Advisor
    Homework Helper

    First off there are two terms in the derivative of the integral since there are two sources of x dependence in the integral (one in the function and one in the limit). The one you have ignored does indeed vanish - is that why you ignored it? Once you get to this:

    dy/dx = -1 + int[dty(t)] from 0 to x

    you have a sign error in the integral. After fixing it you will have a correct equation, but it's still an integral equation. Suggests that you might want to differentiate again, yes?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook