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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?

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$f: x \mapsto 0$ is an elementary solution in $C^2$. – Hans Engler Dec 5 '12 at 23:06

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".

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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.

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In the beginning, did you replace $x$ with $x^2$? – Antonio Vargas Dec 6 '12 at 2:32
The question is about solutions on $(0,\infty)$. There seem to be nontrivial solutions there. – Robert Israel Dec 6 '12 at 3:03

Integrate from $1$ to $x$: the functional equation becomes $$ \int_1^x f''(t)\ dt = \int_1^x f(\sqrt{t})\ dt $$ $$ f'(x) = f'(1) + \int_1^{\sqrt{x}} 2 u f(u)\ du $$ Integrate again: $$ f(x) = f(1) + f'(1) x + \int_1^x \int_1^{\sqrt{t}} 2 u f(u)\ du \ dt = f(1) + f'(1) x + \int_1^{\sqrt{x}} 2 (x - u^2) u f(u)\ du$$ and I think we should be able to use the contraction mapping theorem to get solutions of this in an appropriate space of functions.

EDIT: There are two linearly independent solutions in power series around $x=1$:

$$ 1+\frac{1}{2} \left( x-1 \right) ^{2}+{\frac {1}{96}}\, \left( x-1 \right) ^ {4}-{\frac {1}{320}}\, \left( x-1 \right) ^{5}+{\frac {61}{46080}}\, \left( x-1 \right) ^{6}-{\frac {863}{1290240}}\, \left( x-1 \right) ^ {7}+{\frac {62581}{165150720}}\, \left( x-1 \right) ^{8}-{\frac { 2767907}{11890851840}}\, \left( x-1 \right) ^{9}+{\frac {192672359}{ 1268357529600}}\, \left( x-1 \right) ^{10} + \ldots $$ and $$ x-1+\frac{1}{12}\, \left( x-1 \right) ^{3}-{\frac {1}{96}}\, \left( x-1 \right) ^{4}+{\frac {7}{1920}}\, \left( x-1 \right) ^{5}-{\frac {73}{ 46080}}\, \left( x-1 \right) ^{6}+{\frac {2087}{2580480}}\, \left( x-1 \right) ^{7}-{\frac {76093}{165150720}}\, \left( x-1 \right) ^{8}+{ \frac {6754511}{23781703680}}\, \left( x-1 \right) ^{9}-{\frac { 706941917}{3805072588800}}\, \left( x-1 \right) ^{10} + \ldots $$

I would expect these series to have radius of convergence $1$, with singularities at $0$, but the solutions should extend analytically to $(0,\infty)$.

EDIT: if $f$ satisfies the equation on $(a,b)$, $0 < a < 1 < b < \infty$, then $f(1) + f'(1) x + \int_1^{\sqrt{x}} 2 (x - u^2) u f(u)\ du$ satisfies the equation on $(a^2, b^2)$, and agrees with $f$ on $(a,b)$. So if there is a solution on a neighbourhood of $1$, there is a solution on $(0,\infty)$.

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