From: taga@news.uni-rostock.de (Dr. Michael Ulm) Subject: Re: Integral equation help! Date: 7 Feb 2001 09:58:57 +0100 Newsgroups: sci.math Summary: Solving a typical integral equation On Wed, 07 Feb 2001 08:00:16 GMT, fcueto@my-deja.com wrote: >In article , > taga@news.uni-rostock.de (Dr. Michael Ulm) wrote: >> On Tue, 06 Feb 2001 07:55:32 GMT, fcueto@my-deja.com wrote: >> >First of all, I don't have any experience with integral equations. I >> >came up with a problem that requires to solve the following equation: >> > >> >(b-a)*g(x)= \int_a^b f(x + y) dy >> > >> >where the function g(x) is known for all reals, is bounded and >> >differentiable. The function f(x) is not known, but is expected to be >> >also well behaved. >> >- Is there any analytical or numerical way to solve this equation? >> >- Could you point to any useful online sites relevant to this problem? >> > --snip-- >> >> So, in general you will get infinitely many solutions which are >> easy to construct >> >> The fun starts if you want f to be bounded. >> > >Thank you very much for your careful explanation. As a matter of fact I >would like to have f to be bounded. >Actually I would like to have f: R \to [0,1] > >Would that complicate very much the problem??? > Yes. For one, you will not always get a solution. This can be seen immediately from your original equation. If f is in [0,1] then int_a^b f(x+y) dy will be in [0, b-a], so this can only be possible if g is always in [0,1]. I also realized, that in my first answer I made things a bit too complicated. So here it is again, a bit simpler. We take the derivative of your equation and obtain (b-a) g'(x) = int_a^b f'(x+y) dy = f(x+b) - f(x+a). This will be our new equation. Without loss of generality, we may assume that a=0. We get f(x+b) = f(x) + b g'(x) or f(x) = f(x-b) + b g'(x-b). Now we fix f in the interval [0,b). Then we can get the solution f in the interval [b, 2 b) from the above equation: f(x) = f(x-b) + b g'(x-b) if x in [b,2b) similarly f(x) = f(x-2 b) + b g'(x-b) + b g'(x-2 b) if x in [2 b, 3 b) and so on. By induction, we get f(x) = f(x-n b) + b sum_{k=1}^n g'(x-k b) if x in [n b, (n+1) b). Note, that the term f(x-n b) takes values in our original interval [0, b). Now, if you want f to be bounded near positive infinity, you need some conditions on g', in particular, sum_{l=0}^{infinity} g'(x + l b) must exist for all x. A similar argument gives you that f is bounded at minus infinity only if sum_{l=0}^{infinity} g'(x - l b) must exist. Both conditions can be written as sum_{l = -infinity}^{infinity} g'(x + l b) must exist. Some more fiddling with the equation for f gives also conditions, when an explicit bound on f is possible. So there exists an f with values in [0,1] if and only if there exists an interval I with length 1, such that the numbers sum_{l=r}^s g'(x + l b) all lie in I for all integers r