Differential inequality and blow-up: explanation of part of a proposition I wanted to understand the end of the proof of this proposition, it is part of this paper: ANALYSIS OF A CONVECTIVE REACTION-DIFFUSION EQUATION II*, by H.A. LEVINE, L. E. PAYNE, P. E. SACKS, and B. STRAUGHAN

It is already proved that the equation, which is $u_t =u_{xx} + \varepsilon(u^m)_x + u
^p$ has a solution on some interval $[0,T_{\max})$, and that it blow-up if time $T_{\max}$ is finite. The purpose of this proposition is to prove that under the given hypothesis, the solution blow-up. The beginning is well understood, the author has multiplied the equation by $\psi^n$, where $\psi$ is an eigenfunction of $u_{xx}$ and is doing calculations to obtain a differential inequality that will blowup. Until (4.9) everything is clear. From there I don't quite understand what is being done, it seems that the author uses the fact that $Q$ is increasing, together with Jensen's inequality to conclude (4.12), from (4.10), but I don't know why $\int_{0}^{L}u^p(x,t)\psi^n(x)dx>s_0$. Also about blow-up, I don't know many results about differential inequalities, can you recommend a good book where I can learn about them? If $Q$ were the identity, the blow-up at the end would be clear, as we would have an inequality of the type $y'\geq y^p$. But with that specific $Q$, how to guarantee the blow-up? I would appreciate any clearer explanation of how to better understand the end of this proposition.
 A: Showing that $\int^L_0 u^p(x,t)\psi^n(x) \mathrm{d}x > s_0$.
Note: I suspect that the authors have made some errors in the constants. I shall instead define
$$A_3 = \left(\int^L_0 \psi^n (x) \mathrm{d}x \right)^{(p-1)/p}. $$
(It looks like $A_2$ but without the factor $\frac{\pi^2 n}{L^2}$.)
Let $$I(t) = \int^L_0 u^p(x,t)\psi^n(x) \mathrm{d}x. $$
Then, (4.9) reads
$$F'(t) \geq Q(I(t)).$$
First, from the comment that follows right after (4.10), it says that we can take $s_0$ to be the largest positive root of $Q$ such that $Q'(s) > 0$ for $s > s_0$. Thus, as long as $I(t) > s_0$ for all $t \geq 0$, then by the increasing property of $Q$, we have that
$$F'(t) \geq Q(I(t)) > Q(s_0) = 0$$
(Recall that $s_0$ is a root for $Q$.)
Now, we will try to show that $I(t) > s_0$.
By (4.6) (but without the factor  $\frac{\pi^2 n}{L^2}$), we have
$$\int^L_0 u(x,t)\psi^n(x)\mathrm{d}x \leq \left( \int^L_0 \psi^n(x) \mathrm{d}x \right)^{(p-1)/p} \left( \int^L_0 u(x,t)^p \psi(x)^n \mathrm{d}x \right)^{1/p} $$
or simply put,
$$I(t) \geq A_3^{-p}F^p(t).  \hspace{10pt} (*) $$
In particular,
$$I(0) \geq A_3^{-p}F^p(0). $$
Assuming that our choice of $C_0$ was instead $A_3 s_0^{1/p}$, then by this same choice as in (4.11), we have
$$I(0) \geq A_3^{-p}F^p(0) > s_0 $$
Since $I(0) > s_0$, then from (4.9) at $t = 0$, we have
$$F'(0) \geq Q(I(0)) > Q(s_0) = 0.$$
Since $F(0) = C_0 > A_3 s_0^{1/p} > 0$, note that at $t = 0$, we have that the value of $F$ increases. If the value of $F$ increases, by $(*)$, it is necessary that $I(t)$ increases, and thus for some small value of $t \geq 0$, it remains true that
$$F'(t) \geq Q(I(t)) \geq Q(I(0)) > 0.$$
By repeating this argument rigorously on the interval $[0,T]$ for any $T > 0$ using a bootstrap argument/continuity method, we can show that $F'(t) > 0$ for all $t \geq 0$ and $I(t) \geq I(0) > s_0$ as required.
At this point, the claim in the paper is obtained by expanding the definition of $I(t)$, i.e,
$$I(t) = \int^L_0 u^p(x,t)\psi^n(x) \mathrm{d}x > s_0.$$
Obtaining (4.12).
Since $Q(.)$ is increasing in its argument, from $(*)$, we have
$$F'(t) \geq Q\left( \frac{1}{A_3^p}F^p(t) \right).$$
(Note the difference in the scaling constant inside the argument for $Q$; I don't think this makes a difference in the conclusion that is to be shown which might be why the authors were not that careful about it.)
Finite-time argument
Since $p > m$, then both exponents in $Q(s)$ as in (4.10) are less than $1$. Since in the argument for $Q$, we have shown above that $s \geq s_0$ and such that $Q'(s) > 0$ for $s > s_0$, from the functional form of $Q$:
$$Q(s) = s - A_1 s^{m/p} - A_2 s^{1/p},$$
this is only possible for $s$ large enough (ie $s$ is the dominating term for this case.)
(One way to see this is that for small values of $s$, exponents smaller than $1$ favors the terms $s^{m/p}$ and $s^{1/p}$; only for $s$ large enough can the graph "curves" back up such that it is now $\geq 0$. You can try to prove this fact rigorously although I suspect that the computation is probably non-trivial.)
Thus, for large values of $s$, we have
$$Q(s) \geq Ds $$
for some constant $0 < D < 1$.
(This can be done by decomposing $s = D s + (1 - D)s$ and obtain $(1-D)s - A_1 s^{m/p} - A_2 s^{1/p} \geq 0$ for a sufficiently large $s$ provided)
This implies that (4.12) is given by
$$F'(t) \gtrsim F(t)^p.$$
Since $p > 1$, the standard finite-time blowup arguments hold (as you have mentioned).
Hope this helps! (I'm not a huge expert at this so there might be minor mistakes etc.)
