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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.

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    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

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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.

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    @AlbertH Thanks for the answer :)2012-06-08