I want to show that the $ (t_- , t_+) $ of $$ y'=\cos y \sqrt{t^2+y^2}, y(0)=\pi$$ is $ (t_- , t_+) =\mathbb R$ Therefore I want to show, that f is Lipschitz.
$$ \bigg|\cos y_1 \sqrt{t^2+y_1^2} - \cos y_2 \sqrt{t^2+y_2^2}\bigg|\leq \bigg|\sqrt{t^2+y_1^2}+\sqrt{t^2+y_1^2} \leq\bigg| \frac{t^2+y_1^2-t^2-y_2^2}{\sqrt{t^2+y_1^2}-\sqrt{t^2+y_1^2}}\bigg| $$ This leads to nothing, because I have "-" at the denominator. Is there an easier way?
On
I would write $$\cos(y_1)\sqrt{t^2+y_1^2}-\cos(y_2)\sqrt{t^2+y_1^2}+\cos(y_2)\sqrt{t^2+y_2^2}-\cos(y_2)\sqrt{t^2+y_1^2}$$ The first summand is $$\sqrt{t^2+y_1^2}(\cos(y_1)-\cos(y_2))=\sqrt{t^2+y_1^2}(y_1-y_2)(-\sin(\xi))$$ by the mean value theorem. And $$\sqrt{t^2+y_1^2}-\sqrt{t^2+y_1^2}=(t^2+y_2^2-t^2-y_1^2)\cdot \frac {1}{2}(t^2+\eta^2)^{-1/2}2\eta$$
The function $$ f: \Bbb R \times (0, \infty) \to \Bbb R \, \\ f(t, y) = \cos y \sqrt{t^2+y^2} $$ is locally Lipschitz continuous with respect to $y$ because the partial derivative $$ \frac{\partial}{\partial y} f(t, y) = -\sin y \sqrt{t^2+y^2} + \cos y \frac{y}{\sqrt{t^2+y^2}} $$ exists and is continuous in $ \Bbb R \times (0, \infty)$. Therefore the initial value problem $$ y' = f(t, y) \, , \quad y(0) = \pi $$ has a unique solution in a neighbourhood of $t=0$, and that solution has a “maximal interval of existence” $(t_- , t_+)$. If $t_+ < \infty$ then $$ \tag{*} \lim_{t \nearrow t_+} |y(t)| = \infty $$ must hold, and similarly for the left boundary. (The solution “blows up” at the boundaries of the existence interval. See for example “1.5 Intervals of Existence” in Christopher P. Grant: Theory of Ordinary Differential Equations.)
Now the initial value problems $$ y' = f(t, y) \, , \quad y(0) = \pi/2 \\ y' = f(t, y) \, , \quad y(0) = 3 \pi/2 $$ have the constant solutions $y_1(t) = \pi/2$ and $y_2(t) = 3\pi/2$, respectively. Because of the local uniqueness of solutions, $y, y_1, y_2$ cannot intersect, so that $$ \pi/2 < y(t) < 3 \pi/2 $$ for $t \in (t_- , t_+)$. In other words, $y$ is bounded and $(*)$ cannot hold.
Consequently, the maximal existence interval of $y$ is $(t_- , t_+) = (-\infty, \infty) = \Bbb R$.
Remark: $f(y, t)$ is not globally Lipschitz continuous with respect to $y$ since the partial derivative $\frac{\partial}{\partial y} f(t, y)$ is not globally bounded. Therefore Lipschitz continuity alone is not sufficient to show that the solution exists on $\Bbb R$.