Solution of a Cauchy ODE

Question

Let $y(t)$ be a solution of a Cauchy Problem: $$\dot{y}=\ln\left(\sqrt{1+y^2}\right)$$ with the initial condition $y(0)=y_0$.

1. Prove that if $y_0>0$, then $y$ is a strictly increasing convex ; if $y_0<0$, then $y$ is strictly increasing and concave
2. Prove that $y$ is globally defined

Could someone please help me in understanding how to solve this kind of problems? I would like to be able to solve the whole class of such problems if possible.

Thanks in advance

Answer 1

HINT

To show $y$ is convex it suffices to prove $y'' > 0$. Note that since $y_0 > 0$, you have $$ y'(0) = \ln \left(\sqrt{1+y_0^2}\right) = \ln(1+\epsilon) > 0 $$ and $$ y''(t) = \frac{dy'(t)}{dt} = ... $$ can you argue the result is always positive?

Answer 2

We have that $\sqrt{1+y_t^2} \geq 1 > 0$ and as the function $x \mapsto \ln(x)$ is (strictly) increasing in $\mathbb{R}_+$. Hence, we have : $y_t' \geq 0$. In fact, this inequality is a strict inequality if $y_0 >0$. Finally, we know that a strictly increasing function is a strictly convex.