Pseudo-arclength continuation

One option for choosing $g(x)$ is to select a hyperplane passing through $X^0$ that is orthogonal to the vector $v_i$:
\begin{displaymath}
g(x) = \langle x-X^0,v_i \rangle \ .
\end{displaymath} (3)

So, the Newton iteration becomes:
\begin{displaymath}
X^{k+1} = X^k - H^{-1}_x(X^k)H(X^k)
\end{displaymath} (4)


\begin{displaymath}
H(X) = \left( \begin{array}{c}
F(X)\\
0
\end{array}\...
... \begin{array}{c}
F_x(X)\\
v_i^T
\end{array}\right)\ .
\end{displaymath} (5)

Then one can prove that the Newton iteration for (2) will converge to a point $x_{i+1}$ on the curve from $X^0$ provided that the stepsize $h$ is sufficiently small and that the curve is regular (rank $F_x(x)=n$). Having found the new point $x_{i+1}$ on the curve we need to compute the tangent vector at that point:
\begin{displaymath}
F_x(x_{i+1})v_{i+1} = 0\ .
\end{displaymath} (6)

Furthermore the direction along the curve must be preserved: $\langle v_i,v_{i+1} \rangle = 1$, so we get the ($n+1$)-dimensional appended system
\begin{displaymath}
\left( \begin{array}{c}
F_x(x_{i+1})\\
v_i^T
\end{a...
...
= \left( \begin{array}{c}
0\\
1
\end{array}\right).
\end{displaymath} (7)

Upon solving this system, $v_{i+1}$ must be normalized.