Global solutions for Cauchy problem

Question

By using Picard-Lindelof theorem, it's easy to prove the local existence and uniqueness of the solution for the following Cauchy problem $$ \left\{ \begin{array} [l]{l}% y'=\sqrt{t^2+y^2},\quad t\in\mathbb{R},\\ y(t_0)=y_0,\quad (t_0,y_0)\in\mathbb{R}^2, \end{array} \right. $$ but I can't establish the globality of the solution.

Here is what I do:

$$\forall t\in \mathbb{R},\ \vert\frac{\partial}{\partial y}f(t,y)\vert=\frac{\vert y\vert}{\sqrt{t^2+y^2}}\leq 1,$$ and so $f$ is (globally) Lipschitzian in $y$ on $\mathbb{R}^2$, and then the existence is global.

Is it correct?

Answer 1

$$ |\sqrt{t^2+y^2} - \sqrt{t^2+z^2}| = \frac{|y^2 - z^2|}{ \sqrt{t^2+y^2} + \sqrt{t^2+z^2}} ≤ \frac{|y+z|}{|y|+|z|} |y-z| ≤ |y-z|$$

so $f(t,y) = \sqrt{t^2+y^2}$ is $y$-Lipschitz uniformly in $t$ with constant $1=:L>0$, and is continuous in $t$. By Picard-Lindelöf, There is $\epsilon$ and a unique solution $y$ for times $[t_0 - \epsilon , t_0 + \epsilon]$. Looking through the proof of Picard-Lindelöf, we see that $\epsilon$ (which is $a$ in the wikipedia proof) is only constrained to be $$\epsilon < 1/L$$ Importantly, it does not depend on $t$. Then we iterate as follows, given an initial condition $y(0)$:

• Get a unique solution $y$ on $[0,\epsilon]$ via P-L theorem above.
• Use $\tilde y(0) = y(\epsilon/2)$ as an initial condition.
• Get a unique solution $\tilde y$ on $[0,\epsilon]$.
• If we consider $z(t) := \tilde y(t - \epsilon/2)$, we see that $z(t)$ exists on the interval $[\epsilon/2 , 3\epsilon/2]$ and solves the ODE $\dot{z} = f(t,z)$.
• On the common time of existence $[\epsilon/2,\epsilon]$ for $y$ and $z$, uniqueness of solutions imply that $y=z$.
• Extend $y$ to a function $y:[0,3\epsilon/2]\to ℝ $ by defining $y(t+\epsilon/2) = z(t+\epsilon/2)$.
• Verify that the extended $y$ is still a solution of the ODE.

It is clear we can continually make steps of size $\epsilon/2$ without bound, both forwards and backwards, and in this way we define a solution $y:ℝ \to ℝ$.