Is the Lambert W function the Newton flow of the exponential function?

Question

Is this right?

The Lambert W function, denoted by $W(z)$, is defined as the inverse function of $f(z) = ze^z$. In other words, if $w = W(z)$, then we have $z = w e^w$.

The continuous Newton's method is a technique for finding the roots of a function by following the flow of its gradient vector field. Specifically, given a function $f(x)$ and an initial point $x_0$, the continuous Newton's method generates a solution curve $x(t)$ that satisfies the differential equation:

$$\frac{dx}{dt} = -\nabla f(x)$$

where $\nabla f(x)$ is the gradient vector of $f$ at $x$.

In the case of the function $f(z) = ze^z$, the gradient vector is given by:

$$\nabla f(z) = (1+z)e^z$$

Therefore, the differential equation for the continuous Newton's method is:

$$\frac{dz}{dt} = -(1+z)e^z$$

which is exactly the derivative of the Lambert W function. Therefore, the solution curve generated by the continuous Newton's method starting at a point $z_0$ is given by:

$$z(t) = W(z_0 e^{-t}).$$

In other words, the Lambert W function is the flow of the exponential function under the continuous Newton's method; or equivalently, the W function is the Newton flow of the exponential function.

Answer 1

This makes no sense in this form.

The solution of $z'(t) = - \nabla f(z(t))$ can only have critical points of $f$ as limits. If $f(z) = ze^z$, this means that solutions converge to $z = -1$, which is easily seen directly. This is not a value of the $W$-function.

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 $t > t_0$.

Answer 2

This answer shows a little more clearly the way $z'(t)$ is factored out of the right-hand-side of the equation. It is an attempt at an elaboration of Hans Englers masterful answer given above.

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

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.