ContinuousScalarNewtonFlow

This answer shows a little more clearly the way $z'(t)$ is factored out of the right-hand-side of the equation.

The derivative $z'(t)$ of the solution $z(t)$ to the Newton flow ODE is not supposed to appear on the right-hand side of the solution so it might appear that the formula $u'(t) = \nabla f(z(t))z'(t)$ is incorrect. However, the given formula is right because $\nabla f(z(t))$ drops from the equation when the value of $z'(t)$ is substituted into the differential equation:

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

Let:

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

Differentiate both sides with respect to $t$:

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

Now, using the given differential equation, substitute the value of $z'(t)$:

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

This simplifies to:

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

since $\nabla f(z(t))$ gets factored out and $f(z(t))=u(t)$.

Now, the solution to this differential equation is given by:

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

And since $u(t) = f(z(t))$, making the substitution we have:

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

This equation represents the scalar case of the continuous Newton method.