Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
6  
$f: x \mapsto 0$ is an elementary solution in $C^2$. –  Hans Engler Dec 5 '12 at 23:06
add comment

3 Answers

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".

share|improve this answer
add comment

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.

share|improve this answer
    
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
add comment

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 $(0,\infty)$.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.