SIMON-MARAIS 2020 B3

A cat is trying to catch a mouse in the nonnegative quadrant

$$N = \{(x_1, x_2) \in \mathbb{R}^2 : x_1, x_2 \geq 0\}.$$

At time $t = 0$ the cat is at $(1, 1)$ and the mouse is at $(0, 0)$. The cat moves with speed $\sqrt{2}$ such that its position $c(t) = (c_1(t), c_2(t))$ is continuous, and differentiable except at finitely many points; while the mouse moves with speed 1 such that its position $m(t) = (m_1(t), m_2(t))$ is also continuous, and differentiable except at finitely many points. Thus

$$c(0) = (1, 1), \quad m(0) = (0, 0);$$

$c(t)$ and $m(t)$ are continuous functions of $t$ such that $c(t), m(t) \in N$ for all $t \geq 0$; the derivatives $c'(t) = (c'_1(t), c'_2(t))$ and $m'(t) = (m'_1(t), m'_2(t))$ each exist for all but finitely many $t$; and

$$(c'_1(t))^2 + (c'_2(t))^2 = 2, \quad (m'_1(t))^2 + (m'_2(t))^2 = 1,$$

whenever the respective derivative exists.

At each time $t$ the cat knows both the mouse’s position $m(t)$ and velocity $m'(t)$. Show that, no matter how the mouse moves, the cat can catch it by time $t = 1$; that is, show that the cat can move such that $c(\tau) = m(\tau)$ for some $\tau \in [0, 1]$.