$y'=\sqrt{1+t^2y^2}$, $y(2)=0$
Justify if you can use the Cauchy-Lipschitz theorem and find the biggest interval on which the unique solution exists.
Here is my attempt:
$t_0=2$ and $z=0$ so we can set $K=[2-a,2+a]\times [-b,b]$
How do I justify that $y$ is Lipschitz on K? I can try showing that its derivative is continuous, but how do I do that?
Then, supposing that we can use the theorem, we need to find the maximum:
$F=\max_{(t,y)\in K}\|f(t,y)\| = \sqrt{1+(2+a)^2b^2} = \sqrt{1+a^2b^2+4ab^2+4b^2}$
And so $g(b)=b/F=\frac{b}{\sqrt{1+a^2b^2+4ab^2+4b^2}}$
And here I'm a bit stuck. What would be the maximum of $g$ on our domain?
Then I will have to do $\alpha=\min (a,b/F)$. To do so, in the example given in class, the prof just solved $\max g(a)=a$. Should I do the same thing here?
As $f(t,y) = \sqrt{1+t^2y^2}$ is $\mathcal C^1$ on $\mathbb R^2$, the ODE $y^\prime(t) = f(t,y(t))$ admits a unique maximal solution with the initial condition $y(2)=0$ according to Picard–Lindelöf theorem.
Suppose that the maximal solution is only defined on a bounded interval $[2, d)$ on the right where $d \gt 2$. For all $2 \lt t \lt d$ we have
$$y^\prime(t) = \sqrt{1+t^2y^2(t)} \le \sqrt{1+d^2y^2(t)}$$ and
$$\int_2^t \frac{y^\prime(u)}{\sqrt{1+d^2y^2(u)}} \ du = F(y(t)) - F(y(2))\le (d-2) $$
where $$F(u) = \int_0^u \frac{du}{\sqrt{1+d^2u^2}}.$$ As $F$ is continuous on $[0,\infty)$ and $\lim\limits_{u \to \infty} F(u) = \infty$, $y(t)$ is bounded on $[2,d)$ and therefore doesn't blow up as $t \to d$.
A contradiction proving that the maximal solution of the ODE is defined on $[0, \infty)$ on the right.
You can use a similar method to study the maximal interval on the left.