Integration of this expression $ ( u(t)^2 )' \leq 2M u(t)^2 $

Question

undoubtedly, this is really simple but I don't know how to do this. I'm still learning math so I would be gratefull if you are compassionate to me :)

How do you compute this expression please :

$$ ( u(t)^2 )' \leq 2M u(t)^2 $$ where $ u \in \mathcal C^1( [a,b[) \ ; \ M \in \mathbb R. $

the result has to be : $$ u(t) ^2 \leq u(a)^2 e^{2M (t - a) } $$

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I have any method for expression where you have directly the derivative in it. I don't even know what has to be the boundary of the integral.

Is there any method ? If I had to do this, I would either proceed like this :

• if we replace u by x, we have :

  $$ ( x^2 )' \leq 2M x^2 \implies x^2 \leq 2/3 M x^3 + cst$$ but that's surely false...

• or

  $$ ( u(t)^2 )' \leq 2M u(t)^2 \implies \frac{ ( u(t)^2 )' } { u(t)^2 } \leq 2M \implies ln(u(t)^2) \leq 2Mt + u(a) ^2 $$

Answer 1

Your last approach is the correct one. You would have to discuss that and why $u(t)^2>0$ so that you can divide, and that without changing the inequality sign. Then you need to apply the fundamental theorem more carefully, it has to be $$ \ln (u(t)^2)-\ln (u(a)^2)\le 2M(t-a), $$ which is essentially what the claim is.

Answer 2

Put $v:= u^2$. Then $v' \le Mv$ on $[a,b]$. Now cosider $f(t):=e^{-Mt}v(t)$ and show that $f' \le 0$ on $[a,b]$. Hence $f(t) \le f(a)$ for all $t \in [a,b]$. Your turn !