derivable functions

*210* · Hodge Conjecture Offline Joined: 07 Jul 2008 Posts: 53

Hello everybody

Find all dérivable function f such as $(f'(x))^2 =f(x)$ for all $x \in R$

**Posted:** Tue Nov 03, 2009 1:14 pm

Dr Sonnhard Graubner

hello, it is clear that $f(x)=0$ is one solution, in the other case we have
$\pm\frac{dy}{\sqrt{y}}=dx$
$\pm2\sqrt{y}=x+C$
squaring both sides we get
$y=\frac{1}{4}x^2+\frac{C^2}{4}+\frac{1}{2}xC$
Sonnhard.

**Posted:** Tue Nov 03, 2009 1:25 pm

jmerry

Write that as $(y')^2=y$ ; away from zero, we can solve and get $y'=\pm\sqrt{y}$ . Since the derivative of a function which is differentiable everywhere can't have a jump discontinuity, we choose one sign and stick to it, except possibly at points where $y=0$ .

Solving the separable equation, the real solutions of $y'=\sqrt{y}$ are $\int \frac{dy}{\sqrt{y}}=\int dx$ , or $x=2\sqrt{y}+c$ . Solving for $y$ , that's $y=\frac14\left(x-c)^2$ for real constants $c$ and $x\ge c$ . Similarly, the solutions of $y'=-\sqrt{y}$ are $y=\frac14(c-x)^2$ for real $c$ and $x<c$ .

The first group of solutions run to $x=\infty$ , while the second run to $x=-\infty$ . The constant zero function is also a solution; to get a solution on all of $\mathbb{R}$ , we splice them as follows:
$f(x)=\begin{cases}\frac14(a-x)^2& x\le a\\ 0& a\le x\le b\\ \frac14(x-b)^2& b\le x\end{cases}$ , where $a,b\in [-\infty,\infty]$ (the extended reals) with $a\le b$ .

**Posted:** Tue Nov 03, 2009 1:40 pm

*210* · Hodge Conjecture Offline Joined: 07 Jul 2008 Posts: 53

Thank you Dr Sonnhard Graubner & jmerry for your precious help

**Posted:** Wed Nov 04, 2009 1:59 am