Name and explanation of a Numerical Analysis method for solving systems of non-linear equations In a non-english textbook of Numerical Analysis there is a method for solving systems of non-linear equations. But not only I can't understand how this method is used but I can't even found the name of it in english so to be able to search for it. The direct translation of its name is "Perturbation of Parameters" but I don't think that this is the name used in english. I have searched in various textbooks in the chapters concerning systems of equations in case I find it with no luck.
I will describe the method as it is presented in the textbook and I hope that someone will recognize it and tell me its name in english. If he can also describe it as to how it is used that would be great.
Method:
We have the system of non-linear equations:
\begin{align*}
&f_{1}(x_{1},...,x_{n})=0\\
&\quad \vdots\\
&f_{n}(x_{1},...,x_{n})=0
\end{align*}
Instead of using these equations we use the following ones:
\begin{align*}
&\sigma_{1}^{(0)}(x_{1},...,x_{n})=0\\
&\quad \vdots \\
&\sigma_{n}^{(0)}(x_{1},...,x_{n})=0
\end{align*}
We must already know one solution of the second system which we must now transform slightly, having as a goal to match the first one.
The transformation of the second set of equations is done in $N$ steps by using:
\begin{equation*}
\sigma_{i}^{(K)}(x_{1},...,x_{n})+\frac{K}{N}\left[f_{1}(x_{1},...,x_{n})-\sigma_{i}^{(K-1)}(x_{1},...,x_{n})\right]
\end{equation*}
Where $i=1,2,...,n, \quad K=1,2,...,N$
The known solution of the second set of equations is:
\begin{equation*}
x_{1}^{(0)},...,x_{n}^{(0)}
\end{equation*}
and is used as the first approach in the first step. 
In the last step $K=N$ and at that point the second set of equations matches the first one.
It seems hard to believe that there won't be someone able to identify it.
 A: Let $\mathbf{F}(\mathbf{x}) = 0$ be the original set of equations. Assume that we could solve $\mathbf{S}(\mathbf{x}) = 0$ easily. Let's constuct a parametrical system of equations
$$
\mathbf{F}(\lambda, \mathbf{x}) = \lambda \mathbf F(\mathbf x) + (1 - \lambda) \mathbf S(\mathbf x) = \mathbf S(\mathbf x) + \lambda (\mathbf F(\mathbf x) - \mathbf S(\mathbf x))= 0.
$$
Solving it with $\lambda = 0$ is the same as solve $\mathbf S(\mathbf x) = 0$, which is simple. Solving it with $\lambda = 1$ is hard, since it is the original problem $\mathbf F(\mathbf x) = 0$.
Let's move from $\lambda = 0$ to $\lambda = 1$ in $N$ steps, with $\lambda_k = \frac{k}{N}, k = 0,1,2, \dots, N$.
The solution for $\mathbf{F}(\lambda_0, \mathbf{x}) = 0$ is already known, it is the solution $\mathbf x^{(0)}$ to $\mathbf S(\mathbf x)= 0$. The solution for $\mathbf{F}(\lambda_1, \mathbf{x}) = 0$ should not be far away from that. Using Newton's method we could find the $\mathbf x^{(1)}$ using $\mathbf{x}^{(0)}$ as the initial guess. 
Repeat the same but for $\mathbf{F}(\lambda_2, \mathbf{x}) = 0$ using the solution $\mathbf x^{(1)}$ as the initial guess. Doing this procedure $N$ times brings us to $\lambda_N = 1$ and we finally solve the initial problem.
