How do I show there are no elementary function solutions for the differential equation $f''(x)=f(\sqrt{x}), x>0$? How do I show there are no elementary function solutions for the differential equation $f''(x)=f(\sqrt{x}), x>0$ in the $C^2(0,\infty)$ space solutions?
 A: Assume that a solution $f$ lies in $C^2(0,\infty)$: then $f''\in C^2(0,\infty)$, so $f\in C^4(0,\infty)$ and so on, so $f\in C^\infty(0,\infty)$. 
By using the power series method, it is easy to prove that there are no non-constant analytic solutions in a right neighbourhood of zero, so all the $C^2$ solutions belong to $C^\infty\setminus C^\omega$, and there are no "elementary" functions that belong to this strange space.
Obviously, this strongly depends on the meaning of "elementary". 
A: Expanding on Jack D'Aurizio's answer,
write the equation as
$f(x^2) = f''(x)$.
If $f(x) = \sum_{n=0}^{\infty} a_n x^n$,
$f(x^2) = \sum_{n=0}^{\infty} a_n x^{2n}$
and $f''(x) = \sum_{n=2}^{\infty} n (n-1)a_n x^{n-2}
= \sum_{n=0}^{\infty} (n+2) (n+1)a_{n+2} x^n
$
so $a_{2n}  = (n+2) (n+1)a_{n+2}$
and $a_{2n+1} = 0$.
Setting $n = 0$,
$a_0 = 2a_2$.
Setting $n = 1$,
$a_2 = 6a+3 = 0$,
so $a+0 = 0$.
Setting $n = 2$,
$a_4 = 12 a_4$, so $a_4 = 0$.
If $n = 2k+1$, $a_{4k+2} = 0$.
In particular $a_6 = 0$.
For $n = 4$, $a_8 = 30a_6 = 0$.
Suppose there is a $n > 4$ for which $a_{2n} \ne 0$.
Let $m$ be the smallest such $n$.
Then, since $2m > m+2 > 6$,
$a_{2m} = (m+2)(m+1)a_{m+2} = 0$.
Therefore there is no such $m$,
and $a_n = 0$ for ann $n$.
Thus the only solution is $f(x) = 0$
if $f(x)$ has a power series expansion.
If $f(x) = a x^b$,
$f''(x) = a b (b-1) x^{b-2}$
and $f(\sqrt{x}) = a x^{b/2}$.
For this to be a solution,
$a = a b(b-1)$ and $b/2 = b-2$
or $b=4$ and $a=0$,
so there is no solution of this form.
There may be a non-zero solution not of these forms,
but I do not know of any.
