Example 2

For the scalar equation $x' = \sqrt{1 - x^2}$ one has $I = \mathbf{R},\ H = ] - 1, +1[$. If one takes $t_0 = x_0 = 0$ the corresponding integral is $\sin t$ on the largest interval containing 0 where the derivative of $\sin t$ is positive, that is to say, on $] - \pi/2, +\pi/2[$; at the endpoints of this interval the integral tends to an endpoint of $H$.