Gronwall inequality for $\frac{d}{dt}u(t) \leq C_1u(t) + C_2\sqrt{u(t)}$

Question

I have the inequality $$\frac{d}{dt}u(t) \leq C_1u(t) + C_2\sqrt{u(t)}$$ for a positive function $u$. Is there a Gronwall inequality that I can use to write $$u(t) \leq C_3u(0)?$$ or something similar. I definitely need something where the right hand side is $u(0) \times \text{something}$.

Answer 1

Íf $C_2 = 0$, the sharp inequality holds $$ u(t) \le u(0) e^{C_1 t} $$ and if $C_1 = 0$, there is the sharp inequality $$ u(t) \le \left(\sqrt{u(0)} + \frac{C_2t}{2} \right)^2 = u(0) + C_2t\sqrt{u(0)} + \frac{C_2^2t^2}{4} $$ so you cannot expect an estimate like "$u(t) \le u(0) \cdot K(t)$" for some function $K(t)$.

In the general case, you can estimate $$ \frac{d}{dt} u \le C_1u + C_2\sqrt{u} \le (C_1 + \delta)u + \frac{C_2^2}{4 \delta} $$ for any $\delta > 0$, leading to the estimate $$ u(t) \le \left( u(0) + \frac{C_2^2}{4 \delta (C_1 + \delta)} \right) e^{(C_1 + \delta)t} - \frac{C_2^2}{4 \delta (C_1 + \delta)} . $$ This does not contain an explicit factor $u(0)$, and this cannot be expected. But you could make this more explicit by minimizing the right hand side wrt. $\delta$.