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

Growth order

  1. Sep 17, 2010 #1
    let be the functions f(x) and g(x) defined by an integral equation

    [tex] g(x)= \int_{0}^{\infty}dy K(yx)f(y)dy [/tex]

    then i want to prove that for example [tex] f(x)= O(x) [/tex]

    then using a change of varialbe yx=t i manage to put

    [tex] g(x) \le \frac{C}{x^{2}} \int_{0}^{\infty}dtK(t)t [/tex]

    if the last integral exists , then a simple condition is that there will be a constant so [tex]x^{2} g(x) \le A [/tex] i assume K(t) is ALWAYS positive on the interval [0,oo)
  2. jcsd
  3. Sep 17, 2010 #2
    It's difficult to decide unless one knows the order of g or K.
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Growth order
  1. Tumor growth (Replies: 1)