Find $f$ such that $\frac{\mathrm d^2}{\mathrm dx^2}f(x)=f\left(\sqrt{x}\right)$.

Question

Which non-constant functions $f$ (if any) satisfy $\dfrac{\mathrm d^2}{\mathrm dx^2}f(x)=f\left(\sqrt{x}\right)$ for $x>0$?

I suspect there is no $f$ which satisfies the differential equation, but I cannot prove this.

Answer 1

I messed up in my first response, but realised this may be a lot easier after all. Look at analytic solutions here and expand in series.

$$f(x) = \sum_{j=0}^{\infty} f_j x^j$$

$$f''(x) = \sum_{j=0}^{\infty} (j+1) (j+2) f_{j+2} x^j$$

Plug in $x^2$ in the latter and equate like coefficients. This gives:

$$(j+1) (j+2) f_{j+2} = f_{2j}$$

and

$$f_{2j+1} = 0$$

Now, you can just read off the coefficients by starting with $j=0$ and going through the even values.

$$f_0 = 1 \cdot 2 \cdot f_2$$

$$f_4 = 3 \cdot 4 \cdot f_4$$ or $$f_4 = 0$$

$$f_8 = 5 \cdot 6 \cdot f_6$$

$$f_{12} = 7 \cdot 8 \cdot f_8$$

Can you see how this can be consistently extended?

Answer 2

shouldn't you integrate twice to answer? making the answer (4/15)x^5/2 + c

http://www.wolframalpha.com/input/?i=d%5E2%2Fdx%5E2+4%2F15x%5E%285%2F2%29