closed-form Newton flow of tanh(ln(1+x^2))

Question

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

\begin{equation} \dot{z} (t) = - \frac{f (z (t))}{\frac{d}{d t} f (z (t))} = - \frac{f (z (t))}{\dot{f} (z (t))} \end{equation}

If we let \begin{equation} b (a) = 2 (1 + a^2) + a^4 \end{equation} then define \begin{equation} g (t, a) = e^t - 1 + \frac{2}{b (a)} \end{equation} and \begin{equation} h (t, a) = \sqrt{e^{2 t} - \frac{a^4 (2 + a^2)^2}{b (a)^2}} \end{equation} then there are 4 solutions of $z (t)$ for the Newton flow given by \begin{equation} z (t,a) = \pm \sqrt{\pm \frac{g (t, a) + h (t, a)}{g (t, a)}} \end{equation}

where

lim_(t→∞)z(t,a)∈{0,±i√2}∀a∈ℂ

and

S(lim_(t→∞)z(t,a))=0

Answer 1

The answer is that the derivative of the coordinate function being solved for never appears on the right hand side. Professor Hans Engler was kind enough to verify my calculations, they really are solutions of the Newton flow and there is no need to call it an almost-Newton flow. I doubt if the equation that I thought was the Newton flow could ever be solved at all.. maybe it can be for a function which is identically 0 everywhere or constant... So it can now be stated that

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 $z(t)$ 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

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

Answer 2

For the motivation of the Newton flow you look at the Newton iteration $$ x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)} $$ and compare it to the forward Euler method with step size $\Delta t=1$ for some ODE $\dot x(t)=h(x(t))$, $$ x_{n+1}=x_n+h(x_n)\Delta t. $$ This then gives $h(x)=-\frac{f(x)}{f'(x)}$.

This is all very crude and similar to the relation of gradient descent and gradient flow.

The resulting plot of curved curves is slightly more esthetic than just simply plotting the direction field of the Newton method. The Newton flow lines have almost nothing to do with the actual Newton iteration sequence. Close to a (simple, regular) root the flow is nearly radial towards it. From any point that is different from the root in the plot resolution, the next Newton iterate will be directly at the root location. Only in a sufficiently powerful zoom-in one will discover that it is indeed distinct from the root in a direction that is quite perpendicular to the radial line of the previous point.

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In the current scalar case, the (trivially) separable equation $\dot x=h(x)$ has the implicit solution $$ t+c=\int\frac{dx}{h(x)}=-\int\frac{f'(x)}{f(x)} dx=\ln|f(x)| $$ or with $x(0)=a$ $$ f(x(t))=f(a)e^{-t} $$

Note that $\tanh(\ln(A))=\frac{A^2-1}{A^2+1}$, so that $$ 1+\frac2{x^2(x^2+2)}=e^t+\frac{2e^t}{a^2(a^2+2)} $$ and onwards via the solution of a bi-quadratic equation.