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### Interesting Differential Equation

```Date: 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....

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/
```
Associated Topics:
College Calculus

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