An application of Gronwall's inequality

Question

I want to apply Gronwall's inequality on the differential inequality$$ y'(t)\leq Cy(t)^b $$ where $C$ is a constant and $b>1$. I was wondering if, in the differential form, I can simply define $\beta(t)=Cy(t)^{b-1}$ and rewrite the previous inequality as$$ y'(t)\leq \beta(t)y(t), $$ since $\beta$ is only required to be real-valued and continuous. Then, I would have that $y(t)$ can be estimated by the solution of the ODE $z'(t)=Cz(t)^b$, which is roughly $z(t)=Ct^\frac{1}{1-b}$. Thus$$ y(t)\leq Ct^\frac{1}{1-b}. $$ Is this correct or is there something I'm not seeing?

PS - I have heard of the Bihari-LaSalle inequality, but I'm not sure whether I need to go there.

Answer 1

As long as $y(t)$ is positive you can divide by $y(t)^b$ and then integrate $$ y(t)^{-b}y'(t)\le C\implies y(0)^{1-b}-y(t)^{1-b}\le (b-1)Ct $$ so that $$ y(t)\le \frac{y(0)}{\Bigl(1-y(0)^{b-1}(b-1)Ct\Bigr)^{1/(b-1)}}. $$

Answer 2

You can apply the inequality with $\beta(t) = Cy(t)^{b-1}$, but your conclusion is incorrect: what it actually tells you is that $y(t)$ can be estimated by the solution of $z'(t) = Cy(t)^{b-1}z(t),$ so you're left with the integral inequality $$y(t) \le y(0)\exp\left(\int_0^t Cy^{b-1}(s) \,\mathrm ds \right),$$ which I doubt is any help.

I would instead follow the path of combining the Grönwall inequality with a local Lipschitz estimate: let $z(t)$ be the solution of $z'=Cz^b$ with initial condition $z(0)=y(0)$ and consider the difference $u=z-y$, which we want to show is non-negative. From the ODE for $z$ and the differential inequality for $y$ we find $$u'(t) \ge C\left( z(t)^b - y(t)^b \right).$$

Suppose that $u(t_1) < 0$ at some time $t_1 > 0,$ and let $t_0$ be the last time at which $u \ge0,$ so that $z(t_0)=y(t_0)$ but $z(t) < y(t)$ on $(t_0,t_1].$ Since $[t_0,t_1]$ is compact and $b\ge1$, there is some constant $M$ such that $z(t)^b-y(t)^b \ge M(z(t)-y(t))$ on this interval, and thus we have $u'(t) \ge CMu(t).$ Applying Grönwall to this inequality on $[t_0,t_1]$ shows $u\ge 0$, a contradiction.