SNewtonFlow

The differential equation for the Newton flow $z (t)$ of $f (t)$ is given by

$$ \dot{z} (t) = - \frac{f (z (t))}{\frac{d}{d z(t)} f (z (t))} = - \frac{f (z (t))}{\dot{f} (z (t))} $$

with initial condition $z (0) = a$ where the derivative in the denominator on the right-hand-side is considered to be already evaluated and the chain-rule does not apply since the derivative is really taken with respect to x(t) rather than t itself. The solution when

$$f (t) = S (t) = \tanh (\ln (1 + t^2))$$

can be determined by defining

$$b (a) = 2 (1 + a^2) + a^4$$

$$g (t, a) = e^t - 1 + \frac{2}{b (a)}$$

and

$$h (t, a) = \sqrt{e^{2 t} - \frac{a^4 (2 + a^2)^2}{b (a)^2}}$$

then it can be shown that there exist 4 solutions of $z (t)$ for the Newton flow of $S (t)$ given by

$$ z (t) = \pm \sqrt{\pm \frac{g (t, a) + h (t, a)}{g (t, a)}} = \pm \sqrt{\pm \frac{e^t - 1 + \frac{2}{b (a)} + \sqrt{e^{2 t} - \frac{a^4 (2 + a^2)^2}{b (a)^2}}}{e^t - 1 + \frac{2}{b (a)}}} $$

The limits of these solutions are roots of the S function, that is

$$\lim_{t \rightarrow \infty} z (t, a) \in \lbrace 0, \pm i \sqrt{2} \rbrace \forall a \in \mathbb{C}$$

so that

$$S (\lim_{t \rightarrow \infty} z (t, a)) = 0 \forall a \in \mathbb{C}$$