Re: Re: How to solve this type of equation in Mathematica?
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- Subject: [mg60045] Re: [mg59999] Re: How to solve this type of equation in Mathematica?
- From: Daniel Lichtblau <danl at wolfram.com>
- Date: Tue, 30 Aug 2005 04:43:03 -0400 (EDT)
- References: <dehk3h$c9t$1@smc.vnet.net> <dep7f7$eur$1@smc.vnet.net> <200508280707.DAA28769@smc.vnet.net>
- Sender: owner-wri-mathgroup at wolfram.com
Jeffrey Lyons wrote:
> On Sat, 27 Aug 2005 08:15:35 +0000 (UTC), robert.dodier at gmail.com
> wrote:
>
>
>>It looks to me like this isn't strong enough
>>to yield a unique solution (or even a solution in
>>a well-known class).
>>
>>If you replace f by A f where A is a constant
>>you get A times \int_P^{\infty} (x + P) f(x) dx.
>>So let f be any function s.t. the integral is defined
>>and not equal to zero. (I believe that's a fairly
>>large class.) If the integral I is not equal to 1,
>>replace f by f/I and that's a solution,
>>if I'm not mistaken.
>>
>>The equation is an example of an integral equation,
>>so you might be able to find more info under that
>>heading. Sorry that I can't be more helpful.
>>
>>Robert Dodier
>
>
> Robert,
> thanks so much for your interest in this problem. Your solution,
> replacing f(x) with f(x)/I if f is not equal to 1, still does not
> prove there is a solution for f(x). For all the solutions of this
> type that I can think of the new f(x) is now a function of both x and
> P, ie f(x, P). For example:
> f(x,P) =0.6666 * x^ (-3) * P^(-1)
> is a solution. What I need is a an f that is only a function of the
> variable "x".
>
> Can it be proven that there is/(is not) a solution to my equation if f
> is only a function of the variable x and not a function of "P"?
>
> Again any assistance would be appreciated.
THis appears to be essentially the same question recently asked and
answered on sci.math.symbolic (again by "not-me at nospam.net", but under a
different name):
http://groups.google.com/group/sci.math.symbolic/browse_frm/thread/afc0d04d6fd606bf/3d35c7d35e7b145a?hl=en#3d35c7d35e7b145a
The upshot: Start with
Integrate[(x+p)*f[x], {x,p,Infinity}]==1
Differentiate with respect to p twice, obtain
3*p*f'[p]+2*f[p]==0
Solve this for f, obtain a function in p and an undetermined constant.
By any number of methods, including plugging into the integral, observe
that no value of the constant will remove the functional dependence on
p. Hence such an f[x] cannot be constructed to be differentiable. A
slightly more involved argument using only one derivative will show that
you cannot do this with a nondifferentiable but integrable function either.
Daniel Lichtblau
Wolfram Research
- References:
- Re: How to solve this type of equation in Mathematica?
- From: Jeffrey Lyons <not-me@nospam.net>
- Re: How to solve this type of equation in Mathematica?