Non existence of continuously differentiable function. Prove that there doesn't exist any function $f: \Bbb {R}\rightarrow\Bbb R $ such that $f(0) =1$  and  $ f'(x) =\int_0^x (f(t))^2  \  dt$
My try:
Clearly $f'(x) \geq 0 $ that implies $ f(x) \geq 1+x$, and similarly one can prove $f(x) \geq 1+p(x)  $   where $p(x) $ is polynomial of arbitrarily large degree.
 A: Clearly $f$ is $C^2$ and $f''=f^2$. Assume that $f$ is defined on $[0,\infty)$, then it's easy to see that $f$ is a $C^2$-diffeomorphism $(0,\infty) \rightarrow (1,\infty)$. Multiplicating the above ODE on $f$ with $f'$ and integrating, one finds that $\frac{f^3}{3}-\frac{f'^2}{2}=\frac{1}{3}$.
Therefore, for any $t \geq 0$, $f'(t) = \sqrt{\frac{2(f(t)^3-1)}{3}}$. Now, let $x > 1$, then $$f^{-1}(x) = \int_1^x{(f^{-1})'(t)dt} = \int_1^x{\frac{dt}{f'(f^{-1}(t))}} = \sqrt{3}\int_1^x{\frac{dt}{\sqrt{2(t^3-1)}}}.$$
This integral is convergent over $[0,\infty)$, therefore $f^{-1}$ is bounded. We get a contradiction.
A: If such $f$ existed, then it would be twice differentiable, since
$$
f'(x)=\int_0^x (f(t))^2\,dt, \tag{1}
$$
and the right hand side is differentiable. Hence $f$ would satisfy the IVP:
$$
f(0)=1,\quad f'(0)=\int_0^0 (f(t))^2\,dt =0, \quad f''(x)=\big(f(x)\big)^2.
$$
Then
$$
f''(x)f'(x)=\big(f(x)\big)^2f'(x) \quad\Longrightarrow\quad
\frac{1}{2}\Big(\big(f'(x)\big)^2\Big)'=\frac{1}{3}\Big(\big(f(x)\big)^3\Big)' \\
\quad\Longrightarrow\quad
\frac{1}{2}\big(f'(x)\big)^2=\frac{1}{3}\big(f(x)\big)^3-\frac{1}{3}
\quad\Longrightarrow\quad f'(x)=\frac{2}{3}\sqrt[3]{\big(f(x)\big)^3-1}
$$
The second "$\Longrightarrow$" since $f(0)=1$ and $f'(0)=0$.
Clearly, $f(0)=1$ implies that $f(x)>0$, positive near $x=0$, and hence
$f'(x)>0$, for all $x>0$, due to (1). Thus $f$ is strictly increasing. If $x_0>0$, then $f(x) \ge f(x_0)=y_0>1$, for all $x\ge x_0$ and
$$
f'(x)=\frac{2}{3}\sqrt{\big(f(x)\big)^3-1}\quad\Longrightarrow\quad
\frac{f'(x)}{\sqrt{\big(f(x)\big)^3-1}}=\frac{2}{3}
\quad\Longrightarrow\quad \int_{x_0}^x \frac{f'(t)\,dt}{\sqrt{\big(f(t)\big)^3-1}}=\frac{2(x-x_0)}{3}
$$
Set
$$
F(y)=\int_{y_0}^{y}\frac{ds}{\sqrt{s^3-1}}<\int_{y_0}^{\infty}\frac{ds}{\sqrt{s^3-1}}=M<\infty.
$$
Hence
$$
M>\int_{f(x_0)}^{f(x)}\frac{ds}{\sqrt{s^3-1}}=\int_{x_0}^x \frac{f'(t)\,dt}{\sqrt{\big(f(t)\big)^3-1}}=\frac{2(x-x_0)}{3}
$$
Contradiction, since the right hand side of the above tends to infinity, as $x\to \infty$.
A: The conditions are equivalent to
$$f^{''}(x) = f^2(x)$$
and $f(0)=1$, $f'(0)=0$.
This Cauchy problem has a (unique) local solution defined on an open interval containing $0$. The problem is that it cannot be defined on all $\mathbb{R}$.
Now
$$f{''}(x) = -U'(f(x))$$  (Newton equation of motion, $U$ , the potential  a smooth function ) implies the conservation of energy
$$\frac{1}{2} (f'(x))^2 + U(f(x)) = \textrm{const}$$
Here we get
$$\frac{1}{2}(f'(x))^2 - \frac{f(x)^3}{3} = c $$
so, using $f(0)=1$, $f'(0)=0$
$$(f'(x))^2 = \frac{2}{3}( f(x)^3 - 1)$$
Recall that $f''(x)=f(x)^2\ge 0$, so $f'(x)$ is increasing. Moreover, $f'(0)=0$, therefore, $f'(x) \le 0$ on $(-\infty, 0]$, and $f'(x)\ge 0$ on $[0, \infty)$.
Therefore, on $[0, \infty)$ we have
$$\frac{df}{dx} = (\frac{2}{3}(f^3(x)-1)^{\frac{1}{2}}$$
or
$$\frac{df}{\sqrt{\frac{2}{3}(f^3-1)}}= dx$$
and, using $f(0) = 1$, we get
$$\int_{1}^{f_0}\frac{df}{\sqrt{\frac{2}{3}(f^3-1)}}  = \int_{0}^{x_0} dx= x_0$$
Now, for $f_0 = \infty$, we get a finite value $x_0$ ( the value of a complete elliptic integral). Conclude that the end of the maximal interval for $f$ is finite.  Notice that $f$ is an even function, same applies for the left end.
$\bf{Added:}$
We have
$$\int_{1}^{\infty} \frac{df}{\sqrt{\frac{2}{3}(f^3-1)}} = \frac{\sqrt{6 π} \cdot \Gamma(7/6)}{\Gamma(2/3)}\equiv 2.97448\ldots= x_{\infty}$$
and the function $f$ is defined on $(-x_{\infty}, x_{\infty})$, with limits $\infty$ at both ends of the interval.
Note that the explicit expression for $f$ involves the inversion of elliptic integrals, and so, elliptic functions. Such a simple problem leads to such complex things...
