ExponentialNewtonFlow

The continuous Newton method, if used to solve the scalar equation $f(z) = y$, considers the differential equation

$$z'(t) = \frac{y - f(z(t))}{\nabla f(z(t))}$$

"in the hope that the limit of $z(t)$ exists", to quote John W. Neuberger, who worked a lot on this method. That limit then should be a zero of $y - f(z)$.

Let

$$u(t) = f(z(t))$$

then

$$u'(t) = \nabla f(z(t))z'(t)$$

and the differential equation may be written as

$$u'(t) = y - u(t)$$

which has the solution

$$u(t) = e^{-t}u(0) + (1-e^{-t})y$$

Therefore we always have

$$z(t) = f^{-1} \left(e^{-t} f(z(0)) + (1-e^{-t}) y\right)$$

for the scalar case of the continuous Newton method.

In the case where $f(z) = ze^z$ and $f^{-1}(u) = W(u)$, the differential equation becomes

$$z' = \frac{y - ze^z}{(1+z)e^z}$$

and the solution is

$$z(t) = W\left(e^{-t}z_0e^{z_0} + (1-e^{-t}) y\right)$$

If $y \ge -e^{-1}$, then the solution is defined for all $t \ge 0$ no matter what $z_0$ is and it converges to $W(y)$. If $y < -e^{-1}$, the solution exists only on a finite $t$-interval $[0,t_0]$ and $z(t_0) = -1$, thus the solution cannot be continued for $t > t_0$.