Gronwall-type inequality with nonlinearity

Question

Let $u: (0,\infty) \times \mathbb R \to \mathbb R$. Suppose that $\int_{\mathbb R} u(t,x) dx \ge 0$ (but not necessarily $u >0$). Let $A:(0,\infty) \to \mathbb R$ with $A \ge 0$. Let $\alpha \ge 0$.

Suppose that we know

$$\frac{d}{dt} \alpha\int_{\mathbb R} u(t,x) dx + A(t) \le \int_{\mathbb R}u^2(t,x) dx.$$

Can we prove a Gronwall-type inequality that implies a bound like $$\alpha\int_{\mathbb R} u(t,x) dx + \int_0^t A(t) dt \le e^t\alpha\int_{\mathbb R}u(0,x) dx$$ (or with something else related to $u(0,x)$ on the right-hand side).

This would be true if there was no square on the right-hand side by the classic Gronwall inequality, but I wonder if we can still say something in this case.

Answer 1

The stated inequality cannot hold: Let $\alpha=1$, take $u(t,x)= U$ if $x\in[0,1]$ and $0$ otherwise, and $A(t) = U^2$. Then all the $d/dt$ are 0, but the desired inequality, $$U + tU^2 \le e^{t}U,$$ fails for $t=\log U$ and $U\to +\infty$.