Solving a non-linear system of equations.

The number of variables is the same as the number of equations.

When I fix a set of variables (say $\vec{y}$) and keep another set free (say $\vec{x}$), the system becomes an under-determined, dense, and linear system of the subset of variables $$A(\vec{y})\vec{x} = \vec{b}(\vec{y})\tag{1}\label{system1}$$

where $A(\vec{y})$ is a dense matrix, and $\vec{b}(\vec{y})$ is a dense vector). Let's call this sub-system $\eqref{system1}$ as system 1.

When I fix $\vec{x}$ and keep $\vec{y}$ free, the system becomes an over-determined, sparse, and non-linear system of the subset of variables $$F(\vec{x}, \vec{y}) = 0\tag{2}\label{system2}$$

The Jacobian $J(\vec{x}, \vec{y})$ is available in a closed form. Let's call this sub-system $\eqref{system2}$ as system 2.

About half of the equations in $\eqref{system2}$ are equality constraints that are linear in terms of $\vec{y}$. Each of the constraints is quite sparse and involves only about 5% of all variables.

Can I solve it with the following?

Algorithm 1

  1. Initialize $\vec{x} = \vec{x}_0$, and $\vec{y} = \vec{y}_0$
  2. Fix $\vec{y}_{n - 1}$, solve $A(\vec{y}_{n-1})\vec{x}_{n} = \vec{b}(\vec{y}_{n-1})$ for one of all the possible $\vec{x}_{n}$ because this system is under-determined.
  3. Fix $\vec{x}_{n - 1}$. Perform one iteration of Newton's method for solving $F(\vec{x}_{n-1}, \vec{y}_{n}) = 0$ for $\vec{y}_{n}$.
  4. If not converged, go to step 2.

Algorithm 2

If I replace step 2 in Algorithm 1 by a Newton's method iteration for solving system 1 $\eqref{system1}$, then I guess the steps become a block Newton's method.


But I don't know if these two algorithms can work because system 1 $\eqref{system1}$ is under-determined and system 2 $\eqref{system2}$ is over-determined.

Can this work?


  • $\begingroup$ How do you choose which variables belong to set $\vec{x}$ and which to $\vec{y}$? $\endgroup$
    – Yuriy S
    Oct 10, 2018 at 15:56
  • $\begingroup$ The system is linear in terms of the variables $\vec{x}$. The remaining variables are $\vec{y}$. $\endgroup$
    – R zu
    Oct 10, 2018 at 16:04
  • $\begingroup$ I see. What I would do at a first glance, is add however many variables from Y to the X subset to make the number of equations match number of variables, and linearize the corresponding equations to get the first approximation. Then use the Newton's method on both subsets as intended. But that's just an idea, I'm not sure if it will work $\endgroup$
    – Yuriy S
    Oct 10, 2018 at 16:07
  • $\begingroup$ Can't linearize (theoretically without numerical methods) around the solution because I don't have a good theoretical approximation of the solution. I think Newton's method is equivalent to linearization at the current approximation of the solution, using the Jacobian. Moving variables from Y to X has several disadvantages in terms of programming. It makes system 1 not linear and not dense anymore. That means higher memory usage and forced to use slower methods. Solving sparse system is much slower than solving a dense system. Speed is crucial as I have to speed up the program by 50 times. $\endgroup$
    – R zu
    Oct 10, 2018 at 16:20
  • $\begingroup$ @YuriyS Maybe. Suppose we linearize the whole thing around the current approximation of the solution. Then the question is equivalent to solving a linear system by block matrix method with the variables split unevenly. If that can work, then the scheme in the question can work. $\endgroup$
    – R zu
    Oct 10, 2018 at 16:44


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