4
$\begingroup$

Assume that we have a function $f:\mathbb{R}^+\rightarrow\mathbb{R}^+$ and $D$ a constant. How can we solve the following equation for $f$:

$$ \int^{b}_{x_2=a} \int^{b}_{x_1=a} \frac{1}{(f(x_1)+f(x_2))^2}d x_1 d x_2 = D \log(\tfrac{b}{a})$$

where $0

Thanks.

  • 0
    Are $a$ and $b$ fixed, of supposed to satisfy some conditions?2012-06-07
  • 0
    They are fixed positive real numbers.2012-06-07
  • 1
    Have you tried letting $F(a,b) = \int_a^b \int_a^b (f(x)+f(y))^{-2} \mathrm dx\, \mathrm dy = D(\log b - \log a)$ and comparing partial deriviatives with respect to $a$ and $b$?2012-06-07
  • 0
    I tried it but it doesn't seem to give a solution. We obtain `$ int^{b}_{x=a} (f(x)+f(b))^{-2} dx = D/ (2 b) $` and `$ int^{b}_{x=a} (f(x)+f(a))^{-2} dx = D/ (2 a) $` if I didn't make a mistake somewhere. However using the same method again does simplify them further.2012-06-07
  • 0
    @korpeo Could you let us know in what context you found this problem?2012-06-07
  • 0
    You can take $D$ to be 1, can't you, by toggling $f$ and $\sqrt Df$.2012-06-07
  • 0
    Actually x1 and x2 are random variables and f is a function of them in a problem I am working on. I assumed them to have a uniform(a,b) distribution. The model I am working on resulted in this equation for which I couldn't find a closed-form solution and I think the details of the model is not necessary to solve the equation.2012-06-07
  • 0
    @korpeo Thanks. The reason I asked the context since sometimes the context can provide some insight into the problem.2012-06-08
  • 0
    Unless I've made an error somewhere, $\frac{\partial^2}{\partial a\partial b} F(a,b) = -2(f(a)+f(b))^{-2}$. But $\frac{\partial^2}{\partial a\partial b} D(\log b - \log a) = 0$, so it seems there is no solution.2012-06-08

1 Answers 1

3

Let's suppose $f$ is solution of the above equation, set $$ F(x_1, x_2) = \frac{1}{(f(x_1)+f(x_2))^2} $$ and $b' > b$. Since $$ D\log(b'/b) = D\log(b'/a) - D\log(b/a) $$ we have $$ \int^{b'}_{b} \int^{b'}_{b} F(x_1, x_2) d x_1 d x_2 = \int^{b'}_{a} \int^{b'}_{a} F(x_1, x_2) d x_1 d x_2 - \int^{b}_{a} \int^b_a F(x_1, x_2) d x_1 d x_2 $$ therefore $$ \int^{b}_{a} \int^{b'}_{b} F(x_1, x_2) d x_1 d x_2 + \int^{b'}_{b} \int^{b}_{a} F(x_1, x_2) d x_1 d x_2 = 0 $$ but $F(x_1, x_2)$ is non-negative so $F(x_1, x_2) = 0$ almost everywhere.

No solution can exist.

  • 0
    It's true if as supposed $F$ is solution of the equation.2012-06-08
  • 0
    You're right. Comment deleted.2012-06-08
  • 0
    Come to think of it, I believe your answer is in some sense the "integral form" of my "differential" comment [here](http://math.stackexchange.com/questions/155324/how-to-find-the-function-f-using-this-equation/155401#comment358350_155324).2012-06-08
  • 0
    This statement $$\int_a^{b'}\int_a^{b'} - \int_a^{b}\int_a^{b} = \int_b^{b'} \int_b^{b'}$$ is false. You are missing out on the off-diagonal rectangular blocks of integration.2012-06-08
  • 0
    @Raul In fact, it is.2012-06-08
  • 0
    @Marvis In general it's false, but if $F$ is solution of the equation, it is true. It comes from the equality involving logs.2012-06-08
  • 0
    @Marvis, now I see that I shouldn't have deleted my comment. It is false that $\int_a^c\int_a^c - \int_a^b\int_a^b = \int_b^c\int_b^c$, but it is true under the assumption that $\int_a^b = D(\log b - \log a)$, because $D(\log c - \log a) - D(\log b - \log a) = D(\log c - \log b)$.2012-06-08
  • 0
    Ok. Nice... :) $ $2012-06-08
  • 0
    @AlbertH +1! This is a really nice neat argument!2012-06-08
  • 0
    @Marvis Thanks :)2012-06-08
  • 0
    @AlbertH Thanks for the answer :)2012-06-08