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Proving a real valued function does not exist

  1. Oct 25, 2014 #1
    1. The problem statement, all variables and given/known data
    Show that if [itex] \lambda > \frac{1}{2} [/itex] there does not exist a real-valued function 'u' such that for all x in the closed interval [itex] 0 <= x <= 1 [/itex], [itex] u(x) = 1 + \lambda\int_{x}^{1} u(y)u(y-x)\,dy [/itex]

    2. Relevant equations


    3. The attempt at a solution
    I started off by assuming that such a function does in fact exist, so:
    [itex] u(x) = 1 + \lambda\int_{x}^{1}u(y)u(y-x)\,dy [/itex]
    [itex] \lambda > \frac{1}{2} [/itex]
    [itex] 0 <= x <= 1[/itex]
    [itex] \mu = \int_{0}^{1} u(x)\,dx [/itex]
    [itex] = \int_{0}^{1}\,dx + \int_{0}^{1}\int_{x}^{1} u(y)u(y-x) dy \hspace{1 mm} dx [/itex]
    Here, I changed the order of integration:
    [itex] = 1 + \lambda\int_{0}^{1}\int_{0}^{y} u(y)u(y-x)dx \hspace{1 mm} dy [/itex]
    [itex] = 1 + \lambda\int_{0}^{1}u(y)\int_{0}^{y}u(y-x)dx \hspace{1 mm} dy [/itex]
    At this point I am a bit confused about where to go, so I am starting to think that perhaps I am not going in the right direction; I was hoping someone could give me some advice. Thanks :)
     
  2. jcsd
  3. Oct 26, 2014 #2
    Ok, I worked on this for a quite a while last night before bed and I think I have a solution, hopefully someone can confirm/deny...I'll start where I left off above, where I take Y constant:
    [itex] z = y - x [/itex]
    [itex] dz = -dx [/itex]
    [itex] \mu = 1 + \lambda\int_{0}^{1}u(y)\int_{y}^{0} u(z)(-dz)\,dy [/itex]
    [itex] = 1 + \lambda\int_{0}^{1}u(y)\int_{0}^{y}u(z)dz\,dy [/itex]
    [itex] {Let} \hspace{1 mm} f(y) = \int_{0}^{y} u(z)\,dz \hspace{2 mm} {then,} \hspace{2 mm} f'(y) = u(y), \hspace{1 mm} f(0)=0, \hspace{1 mm} f(1) = \mu [/itex]
    [itex] \mu = 1 + \lambda\int_{0}^{1}f'(y)f(y)\,dy [/itex]
    [itex] = 1 + \frac{\lambda {\mu}^2}{2} [/itex]
    [itex] \lambda {\mu}^2 - 2\mu + 2 = 0 [/itex]
    So now, if mu is going to exist, the discriminant of that quadratic cannot be negative:
    [itex] 4 - 8\lambda >= 0 [/itex]
    [itex] \lambda <= \frac{1}{2} [/itex]
    -This contradicts the problem statement that [itex] \lambda > \frac{1}{2} [/itex]. Hence, it is confirmed that for [itex] \lambda > \frac{1}{2} [/itex] there does not exist [itex] u = 1 + \lambda\int_{x}^{1}u(y)u(y-x)\,dy[/itex].
     
  4. Oct 26, 2014 #3

    pasmith

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    Homework Helper

    Looks correct.
     
  5. Oct 26, 2014 #4
    thanks :)
     
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