From: chrisman@ucdmath.ucdavis.edu (Mark Chrisman) Newsgroups: sci.math Subject: An integral equation... Date: Tue, 31 Oct 1995 19:06:23 -0800 Ok, try this out. If the real function F is defined on [0,1], find G such that a 1 F(a) = Integral Integral G(x,y) dy dx. 0 a (For what functions F can we find such a G? Is G easily computable from F?) -Mark -- Mark Chrisman chrisman@ucdmath.ucdavis.edu ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: An integral equation... Date: 2 Nov 1995 07:21:38 GMT In article , Mark Chrisman wrote: >Ok, try this out. What was that thread about mathematics not being experimental? About how it's all logical deductions from axioms? Here we have a problem which is wonderfully vague. We can go about answering it for a long time, forever changing the precise conditions under which the problem is to be solved, looking for the most elegant or useful answer. I'll take the easy way out, and assume it's OK if G is just integrable. >If the real function F is defined on [0,1], find G such that > > a 1 >F(a) = Integral Integral G(x,y) dy dx. > 0 a Let S_a be the rectangle [0,a]x[a,1] which is just above the main diagonal in the square. The poster assumes given an F: [0,1] -> R and asks if a G exists with F(a)=integral_(S_a) G, presumably asking as well if G must be unique, and enquiring as to the computability of G. First off, note that uniqueness is out. Among other things, the condition which is to determine G makes no reference to the values of G below the main diagonal, so if any solution exists, there are many; in fact there are many in just about any function-space in which there is one, except possibly C^\omega([0,1]^2). (There's a lot more freedom too, as we shall see.) Well, if the value of G doesn't matter down there, we can simply redefine G(x,y)=-G(y,x) for x > y. This has the nice effect of making integral_S G = 0 for any set S which is symmetric about the main diagonal, so that in particular the integral over [0,a]x[0,a] is zero, and thus integral_(S_a) G is the same as the integral of G over [0,a] x [0,1]. Assuming G is nice enough that we can Fubinate, this integral is integral_[0,a] g(x) dx, where for each x in [0,1] g(x) = integral_[0,1] G(x,y) dy............(*) So now the question is, what are the skew-symmetric functions G for which the corresponding g has F(a)=integral_[0,a] g for all a? Now, the poster didn't specify what sort of function F is, but if you hope to be able to find a G, this F had better be differentiable by the Fundamental Theorem of Calculus, and indeed F'(a)=g(a) [almost everywhere]. Moreover, we have recast the problem as: given g:[0,1]->R, find a skew symmetric G satisfying (*). One solution to this is easy. Given g, divide the unit square into four quadrants. Define G to be identically zero in the upper right quadrant. On the lower right quadrant, define G(x,y)=2*g(x); then skew-symmetry requires G(x,y)=-2*g(y) in the upper left quadrant. Let k = -2*integral_[0.5, 1] g(y) dy. In the lower left quadrant we need to find a skew-symmetric function G with the property that for each x in [0,0.5] we have integral_[0,0.5] G(x,y) dy = g(x)-k Clearly, this problem is isomorphic to the one we faced on the whole square! Thus we can iterate this procedure, finding after n passes that we have defined G on all of the square except on [0,1/2^n]^2; in the limit, G is defined everywhere except at (0,0). Along the way, we have discovered there are many choices for G. Indeed, we could have chosen an arbitrary skew-symmetric G in the upper-right quadrant to begin with, adjusted g corresponding, then filled in the next two quadrants, and iterated as before. Roughly speaking, the situation is that the condition that G be skew-symmetric hardly effects the size of the solution set at all; we're still trying to pick G simply knowing its column means, which has an awful lot of degrees of freedom. dave ============================================================================== From: rusin@washington.math.niu.edu (Dave Rusin) Newsgroups: sci.math Subject: Re: An integral equation... Date: 2 Nov 1995 07:54:56 GMT In article , Mark Chrisman wrote: >If the real function F is defined on [0,1], find G such that > > a 1 >F(a) = Integral Integral G(x,y) dy dx. > 0 a If you look for a function G which is constant on horizontal lines, you find the unique solution G(x,y) = (F(y)/y)'. dave