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Problem

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.

Question

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?

Related

Problem

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.

Question

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?

Related

Problem

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.

Question

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?

Related

added 308 characters in body
Source Link
R zu
  • 163
  • 9

Problem

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.

Question

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?

Related

Problem

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.

Question

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?

Related

Problem

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.

Question

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?

Related

added 308 characters in body
Source Link
R zu
  • 163
  • 9

Problem

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.

Question

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?

Related

Problem

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.

Question

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?

Related

Problem

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.

Question

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?

Related

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