Interesting Differential EquationDate: 06/18/2004 at 11:14:00 From: John Subject: whether a rather unusual differential eq. has a solution Perhaps you will consider a question from one who has been out of college for many years (I was a math and physics major). I heard this question posed many years ago but have not been able to find an answer. The problem is: Is there a nontrivial function f(x) such that df/dx = f(f(x)) ? If there is, how would one solve for or construct it? What would it look like? Perhaps a power series solution (though that gets very involved), or a numerical solution with a finite difference method starting from some initial guess and iterating? Date: 06/18/2004 at 21:41:47 From: Doctor Vogler Subject: Re: whether a rather unusual differential eq. has a solution Hi John, That's a very challenging problem, there. But I can give you one way to get a power series solution. If you keep differentiating your equation, you can get all of the derivatives in terms of iterating the function f. For example, f'(x) = f(f(x)) f''(x) = f'(f(x))f'(x) = f(f(f(x)))f(f(x)) f'''(x) = f'(f(f(x)))f'(f(x))f'(x)f(f(x)) + f(f(f(x)))f'(f(x))f'(x) = f(f(f(f(x))))f(f(f(x)))f(f(x))f(f(x)) + f(f(f(x)))f(f(f(x)))f(f(x)) etc. Now, many functions have fixed points. Suppose that f(a) = a for some a. Then you can solve for *all* of the derivatives at a. For suitable values of a, this will probably give you a power series for f(x) in powers of x-a which converges in some region around a. The easy case is a = 0, but then that gives you the trivial solution f(x) = 0 which you didn't want, so use some other value of a. And of course there may be other solutions that don't have a fixed point.... If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. I'll leave the question on the floor in case some other doctor has some other ideas. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ Date: 06/19/2004 at 15:54:56 From: Doctor Mitteldorf Subject: Re: whether a rather unusual differential eq. has a solution Dear John - In answering students' questions, I usually try to stimulate their own thinking and create a dialogue rather than to provide complete answers. Even though you're older, I'll do the same with you. Here's a path you might enjoy going down; try a solution of the form f(x) = ax^n. Then f'(x) = na x^(n-1) f(f(x)) = a (ax^n)^n = a^(n+1) x^(n^2) Identifying the two expressions yields a quadratic equation for n and an exponential expression for a in terms of n, which can be solved numerically. I believe this will lead to a solution of your differential equation, albeit a complex function of a complex variable. For further exploration, it might be fun to look explicitly at the real and imaginary parts of the function. Plot them over the complex plane. See if the form of the solution suggests a way to solve the differential equation on the real line. Will you follow up and let me know what you find? - Doctor Mitteldorf, The Math Forum http://mathforum.org/dr.math/ |
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