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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 $0$.

fedja
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  • Is there one example of $f$ for which you can answer the question? – Mikael de la Salle Jun 20 '18 at 11:26
  • @MikaeldelaSalle Yes. Pure power and exponents are OK (because there is a self-similarity, so once you know one solution, you can describe the whole family and you can see explicitly where you hit the $x$-axis when starting from $(0,u_0)$). – fedja Jun 20 '18 at 13:28
  • I've been thinking about this on and off for the last week. Do you think that you really need $f(u)$ to be greater than all linear functions? If $f(u) = Bu$, then the solution is $u(x) = C J_0(2 \sqrt{Bx})$, so it seems it should be enough to have $f(u)$ grow faster than $B_0 u$, where $B_0$ is chosen such that $J_0(2 \sqrt{B_0}) < 0$. – David E Speyer Jan 08 '19 at 18:10
  • @DavidESpeyer I'm not sure that $f(u)$ being eventually larger than $B_0u$ would be enough. Say that $f(t)=Bt$ for $0<c\le t\le u_0$, with $B>>B_0$, and that the solution $u(x)$ coincides with $J_0(2\sqrt{Bx})$ for $0< x < u^{-1}(c )$. Nevertheless when we take $x > u^{-1}(c )$ in the equation for $\dot u(x)$, we average the values of $f(u(x))$ where it is possibly small, so that the derivative of $u(x)$ need not to be as negative as that of $J_0(2\sqrt{Bx})$ (so $u(x)$ may stay positive and land at $x>1$ without crashing). – Pietro Majer Apr 30 '19 at 19:07

1 Answers1

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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.

Pietro Majer
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