An upper bound for the solution of an integro-differential equation For those who are interested in "motivations" this has something to do with modeling flames in turbulent jets. However the question itself is irritatingly elementary and requires no mathematical or applied background whatsoever. Here is the simplest possible version:
Assume that $f$ is a non-negative increasing convex function on $\mathbb R$ that grows faster than any linear function at $+\infty$.
Suppose that there exists a positive differentiable on $[0,1]$ solution of the IVP $u(0)=u_0\in \mathbb(0,+\infty)$, $u'(x)=-\frac 1x\int_0^x f(u(s))\,ds$ (at $x=0$ the RHS is understood as $-f(u_0)$). Can we show that $u_0$ cannot be arbitrarily large (the exact bound may depend on $f$, of course)?
Note that the finiteness of $u_0$ is assumed a priori and is essential. There exist examples where the solution stays finite positive on $(0,1]$ but blows up at $0$.  
 A: Here are some observations, and an experiment of a bound that works at least for some $f$.
Let $u(x)$ a positive bounded solution as stated in the OP. Differentiating the equation $\dot u(x)=-{1\over x}\int_0^xf(u(s))ds$  we also have
$$\ddot u(x)={1\over x^2}\int_0^x\big[f(u(s))-f(u(x))\big]ds$$
so $u(x)$ is positive, decreasing, and convex on $[0,1]$. 
Therefore $f(u(t))$ is also positive, decreasing, and convex, and since its derivative at $x=0$ is $f'(u(0))\dot u(0)=-f'(u_0)f(u_0)$ we have for $0\le t \le 1$
$$f(u(t))\ge \big(f(u_0)- f'(u_0)f(u_0)t\big)_+= f(u_0)\big(1-f'(u_0)t\big)_+.$$
On the other hand, if we integrate the expression for $\dot u(s)$:
$$u(0)=u(1)-\int_0^1\dot u (s)ds=u(1)+\int_0^1{1\over s}\int_0^s  f(u(t))dtds$$
$$=u(1)+\int_0^1\int_t^1 {ds\over s} f(u(t))dt=u(1)+\int_0^1\log(1/t) f(u(t))dt.$$
Here, plugging the inequality for $f(u(t))$
$$u_0\ge f(u_0) \int_0^{1/f'(u_0)}\log(1/t)\big(1-f'(u_0)t\big) dt={ 3+2\log f'(u_0)\over4f'(u_0) }f(u_0).$$
The latter implies a bound on $u_0$ provided $$\liminf_{x\to+\infty}{ \log f'(x)\over xf'(x)}f(x)>2,$$  which is the case e.g. for $f\sim x^p$ for $p>1$ and $f\sim e^{cx^\alpha}$ for $0<\alpha <1/2$, but fails for $f=e^x$. Improving the above lower bound on $f(u(x))$ should hopefully allow to extend the argument.
