4
$\begingroup$

Pardon my lack of tex skills, but what is the recommended procedure in the following scenario:

$$g(f) = 1+\int_0^{1-f} g\left(\dfrac{f}{1-x}\right)\,dx$$

I am not sure how to proceed in such a scenario. My expression is more complicated, but that is the gist of the concept I'm struggling with.

also, we know that g(1) = 1

I'm thinking some sort of Leibniz approach, but I'm an engineer by training so I'm out of my depth.

edit: If the above simplification does not have a solution/doesn't lend itself well to an example, here is the actual thing:

$$g(f) = [1-(1-f)^{2}] + 2\int_0^{1-f} (1+g\left(\dfrac{f}{1-x}\right))x\,dx$$

  • 0
    I think I've $\LaTeX{}$'d this up correctly - if not then feel free to revert the edit!2012-09-20
  • 0
    that is perfect. thank you!2012-09-20
  • 0
    If you take the derivative on both sides with respect to $x$ you get an intresting property of $g$.2012-09-20
  • 0
    the fact that there is an initial condition suggested this is the approach, but I just cannot make it work2012-09-20
  • 0
    Ah but i see you edited. Hmm. (Thinking)2012-09-20
  • 0
    If we diff with respect to x : 0 = d1 + g(f/(1-(1-f))) - g(f/(1-0)) thus 0 = d1 + g(1) - g(f) Thus it appears that g(f) is a constant function assuming I did not make a mistake.2012-09-20
  • 0
    A friend suggested that the correct approach is differentiating the function thrice. However, I gave that an elementary stab using the Leibniz rule definitions I found online and I couldn't get rid of the integral itself. The solution should most likely be a differential equation in the form g'''(f) = a*g''(f) + b*g'(f) + c*g(f), which I should hopefully then be able to solve into g(f)2012-09-20
  • 0
    where a, b and c are all functions of f2012-09-20
  • 0
    What did you think about my comment ? Did I make a mistake ?2012-09-20
  • 0
    well, the more I looked at the "simplified" example I gave the more I realized that it just wouldn't do. The itnegral from 0 to 0 would be 0 and then g(1) would = 0 instead of the given starting condition of 1. The answer is most definitely not constant. I suspect it is a differential equation of the form above.2012-09-20
  • 0
    I would appreciate if you ( or someone else ) could explain my mistakes then.2012-09-20
  • 0
    @mick I cannot make sense of your approach. Neither side is a function of $x$, so the $x$-derivative of both sides is $0$.2012-09-20
  • 0
    I assumed differentiating dx removed the integral on the RHS ?2012-09-20
  • 0
    @mick Compare what you did to the Fundamental Theorem of Calculus. There are gaping discrepancies.2012-09-20
  • 2
    I see now how wrong i was. Feel silly :)2012-09-21

1 Answers 1

1

Here's a solution to the simplified version. There is a large amount of miraculous cancellation so maybe someone can find a more elegant way to attack it.

The derivative of the RHS (with respect to $f$) has two parts (you could view this as an application of the multivariable chain rule):

$$\frac{d}{df} \int_0^{1-f} g\left(\dfrac{f}{1-x}\right)\,dx = -g\left(\frac{f}{1-(1-f)}\right) + \int_0^{1-f} g'\left(\frac{f}{1-x}\right) \frac{1}{(1-x)}\,dx$$ $$= -1 + \int_0^{1-f} g'\left(\frac{f}{1-x}\right) \frac{1}{(1-x)}\,dx. $$

One can then apply integration by parts to this new integral using the fact that $$ \frac{d}{dx} g\left(\frac{f}{1-x}\right) = g'\left(\frac{f}{1-x}\right) \frac{f}{(1-x)^2}.$$ This gives $$\int_0^{1-f} g'\left(\frac{f}{1-x}\right) \frac{f(1-x)}{f(1-x)^2}\,dx = \frac{1-x}{f} g\left(\frac{f}{1-x}\right) \Bigg|_{x=0}^{1-f} + \int_0^{1-f} \frac{1}{f} g\left(\frac{f}{1-x}\right)\, dx $$ $$ = 1 - \frac{g(f)}{f} + \frac{1}{f}\left(g(f)-1\right) = 1 - \frac1f.$$

This means that the derivative of the RHS (which is also equal to $g'(f)$) is just $-1/f$, so $g(f) = 1 - \ln f$. This does seem to satisfy the original (simplified) equation.

  • 0
    this looks like the right direction I think the miraculous cancellations are a feature, not a bug haha2012-09-20
  • 0
    Im a bit confused.2012-09-20