2
$\begingroup$

I have the following differential inequality: $$f'(x)\geq cf(x)^a,\quad \forall x\in[0,1]$$ where $0<a<1,\, f(x)\geq 0$.


I'm taking the following approach to solve the problem: $$f'(x)\geq cf(x)^a\Rightarrow \frac{df}{dx}f(x)^{-a}\geq c$$ Integrating from $\alpha$ to $t$ gives: $$\int_{x=\alpha}^t f(x)^{-a}df-c\, dx\geq 0$$ Then: $$\frac{f(t)^{1-a}}{1-a}-\frac{f(\alpha)^{1-a}}{1-a}\geq c(t-\alpha)$$ $$ f(t)\geq \left(c(1-a)t+f(\alpha)^{1-a}-c\frac{\alpha}{1-a}\right)^\frac{1}{1-a}$$

This answer seems different from the one obtain in my previous question. Is this answer correct?

Also there exist some sort of initial condition in this solution. I mean to obtain a lower bound on $f(x)$ one should know the value of $f$ at some point $\alpha$. But the derivation seems to be true for all $\alpha$-s. Hence I conclude:

$$f(t)\geq \left(c(1-a)t+f(\alpha)^{1-a}-c\frac{\alpha}{1-a}\right)^\frac{1}{1-a} \forall \alpha\in[0,1]$$ $$f(t)\geq \max_{\alpha\in[0,1]}\left(c(1-a)t+f(\alpha)^{1-a}-c\frac{\alpha}{1-a}\right)^\frac{1}{1-a}$$

If this conclusion is true there doesn't exist any closed form lower-bound on $f(x)$ given initial condition for a single point (since the lower-bound depends on $f$ over the interval $[0,1]$). With this respect I think I didn't actually solve the problem. Because the solution to $f$ depends on $f$ over the entire interval. so

Is it possible to find a closed form necessary and sufficient condition on $f(x)$ to be a solution of $f'(x)\geq cf(x)^a$?

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.