Solving a system of semilinear equations: Will this method work?











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I understand the steps in solving a system of equations



$$mathbf{A}(mathbf{x})mathbf{x} = mathbf{b}$$



Where $mathbf{A}(mathbf{x})$ is a matrix that depends on the solution vector $mathbf{x}$



i) Start with an initial guess for $mathbf{x}$, $mathbf{x_0}$



ii) Compute the residual $mathbf{r} = mathbf{A}(mathbf{x})mathbf{x} - mathbf{b}$



iii) Repeat the following three steps until convergence is obtained




  • Solve $mathbf{A}(mathbf{x})mathbf{w} = mathbf{b}$

  • Set $mathbf{x}$ to $mathbf{w}$

  • Compute the residual $mathbf{r} = mathbf{A}(mathbf{x})mathbf{x} - mathbf{b}$


My question: Can I adapt these steps for a system of semilinear equations



$$mathbf{A}mathbf{x} = mathbf{b}(mathbf{x})$$



I.e.



i) Start with an initial guess for $mathbf{x}$, $mathbf{x_0}$



ii) Compute the residual $mathbf{r} = mathbf{A}mathbf{x} - mathbf{b}(mathbf{x})$



iii) Repeat the following three steps until convergence is obtained




  • Solve $mathbf{A}mathbf{w} = mathbf{b}(mathbf{x})$

  • Set $mathbf{x}$ to $mathbf{w}$

  • Compute the residual $mathbf{r} = mathbf{A}mathbf{x} - mathbf{b}(mathbf{x})$










share|cite|improve this question


























    up vote
    0
    down vote

    favorite












    I understand the steps in solving a system of equations



    $$mathbf{A}(mathbf{x})mathbf{x} = mathbf{b}$$



    Where $mathbf{A}(mathbf{x})$ is a matrix that depends on the solution vector $mathbf{x}$



    i) Start with an initial guess for $mathbf{x}$, $mathbf{x_0}$



    ii) Compute the residual $mathbf{r} = mathbf{A}(mathbf{x})mathbf{x} - mathbf{b}$



    iii) Repeat the following three steps until convergence is obtained




    • Solve $mathbf{A}(mathbf{x})mathbf{w} = mathbf{b}$

    • Set $mathbf{x}$ to $mathbf{w}$

    • Compute the residual $mathbf{r} = mathbf{A}(mathbf{x})mathbf{x} - mathbf{b}$


    My question: Can I adapt these steps for a system of semilinear equations



    $$mathbf{A}mathbf{x} = mathbf{b}(mathbf{x})$$



    I.e.



    i) Start with an initial guess for $mathbf{x}$, $mathbf{x_0}$



    ii) Compute the residual $mathbf{r} = mathbf{A}mathbf{x} - mathbf{b}(mathbf{x})$



    iii) Repeat the following three steps until convergence is obtained




    • Solve $mathbf{A}mathbf{w} = mathbf{b}(mathbf{x})$

    • Set $mathbf{x}$ to $mathbf{w}$

    • Compute the residual $mathbf{r} = mathbf{A}mathbf{x} - mathbf{b}(mathbf{x})$










    share|cite|improve this question
























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      I understand the steps in solving a system of equations



      $$mathbf{A}(mathbf{x})mathbf{x} = mathbf{b}$$



      Where $mathbf{A}(mathbf{x})$ is a matrix that depends on the solution vector $mathbf{x}$



      i) Start with an initial guess for $mathbf{x}$, $mathbf{x_0}$



      ii) Compute the residual $mathbf{r} = mathbf{A}(mathbf{x})mathbf{x} - mathbf{b}$



      iii) Repeat the following three steps until convergence is obtained




      • Solve $mathbf{A}(mathbf{x})mathbf{w} = mathbf{b}$

      • Set $mathbf{x}$ to $mathbf{w}$

      • Compute the residual $mathbf{r} = mathbf{A}(mathbf{x})mathbf{x} - mathbf{b}$


      My question: Can I adapt these steps for a system of semilinear equations



      $$mathbf{A}mathbf{x} = mathbf{b}(mathbf{x})$$



      I.e.



      i) Start with an initial guess for $mathbf{x}$, $mathbf{x_0}$



      ii) Compute the residual $mathbf{r} = mathbf{A}mathbf{x} - mathbf{b}(mathbf{x})$



      iii) Repeat the following three steps until convergence is obtained




      • Solve $mathbf{A}mathbf{w} = mathbf{b}(mathbf{x})$

      • Set $mathbf{x}$ to $mathbf{w}$

      • Compute the residual $mathbf{r} = mathbf{A}mathbf{x} - mathbf{b}(mathbf{x})$










      share|cite|improve this question













      I understand the steps in solving a system of equations



      $$mathbf{A}(mathbf{x})mathbf{x} = mathbf{b}$$



      Where $mathbf{A}(mathbf{x})$ is a matrix that depends on the solution vector $mathbf{x}$



      i) Start with an initial guess for $mathbf{x}$, $mathbf{x_0}$



      ii) Compute the residual $mathbf{r} = mathbf{A}(mathbf{x})mathbf{x} - mathbf{b}$



      iii) Repeat the following three steps until convergence is obtained




      • Solve $mathbf{A}(mathbf{x})mathbf{w} = mathbf{b}$

      • Set $mathbf{x}$ to $mathbf{w}$

      • Compute the residual $mathbf{r} = mathbf{A}(mathbf{x})mathbf{x} - mathbf{b}$


      My question: Can I adapt these steps for a system of semilinear equations



      $$mathbf{A}mathbf{x} = mathbf{b}(mathbf{x})$$



      I.e.



      i) Start with an initial guess for $mathbf{x}$, $mathbf{x_0}$



      ii) Compute the residual $mathbf{r} = mathbf{A}mathbf{x} - mathbf{b}(mathbf{x})$



      iii) Repeat the following three steps until convergence is obtained




      • Solve $mathbf{A}mathbf{w} = mathbf{b}(mathbf{x})$

      • Set $mathbf{x}$ to $mathbf{w}$

      • Compute the residual $mathbf{r} = mathbf{A}mathbf{x} - mathbf{b}(mathbf{x})$







      systems-of-equations nonlinear-system






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      asked Nov 16 at 13:11









      DJames

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          It may work. I assume $A$ is nonsingular. Suppose the true solution is $x_0$ and you are at $x = x_0 + y$ with $y$ small. Let $J$ be the Jacobian of $b$ at $x = x_0$.

          Then you get $w = A^{-1} b(x_0 + y) approx A^{-1}(b(x_0) + J y) = x_0 + A^{-1} J y$. If all eigenvalues of $A^{-1} J$ have absolute value less than $1$, you will converge to the solution $x_0$ provided you start close enough. If there is an eigenvalue with absolute value $> 1$, you will almost certainly not converge to $x_0$.






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            It may work. I assume $A$ is nonsingular. Suppose the true solution is $x_0$ and you are at $x = x_0 + y$ with $y$ small. Let $J$ be the Jacobian of $b$ at $x = x_0$.

            Then you get $w = A^{-1} b(x_0 + y) approx A^{-1}(b(x_0) + J y) = x_0 + A^{-1} J y$. If all eigenvalues of $A^{-1} J$ have absolute value less than $1$, you will converge to the solution $x_0$ provided you start close enough. If there is an eigenvalue with absolute value $> 1$, you will almost certainly not converge to $x_0$.






            share|cite|improve this answer

























              up vote
              0
              down vote













              It may work. I assume $A$ is nonsingular. Suppose the true solution is $x_0$ and you are at $x = x_0 + y$ with $y$ small. Let $J$ be the Jacobian of $b$ at $x = x_0$.

              Then you get $w = A^{-1} b(x_0 + y) approx A^{-1}(b(x_0) + J y) = x_0 + A^{-1} J y$. If all eigenvalues of $A^{-1} J$ have absolute value less than $1$, you will converge to the solution $x_0$ provided you start close enough. If there is an eigenvalue with absolute value $> 1$, you will almost certainly not converge to $x_0$.






              share|cite|improve this answer























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                up vote
                0
                down vote









                It may work. I assume $A$ is nonsingular. Suppose the true solution is $x_0$ and you are at $x = x_0 + y$ with $y$ small. Let $J$ be the Jacobian of $b$ at $x = x_0$.

                Then you get $w = A^{-1} b(x_0 + y) approx A^{-1}(b(x_0) + J y) = x_0 + A^{-1} J y$. If all eigenvalues of $A^{-1} J$ have absolute value less than $1$, you will converge to the solution $x_0$ provided you start close enough. If there is an eigenvalue with absolute value $> 1$, you will almost certainly not converge to $x_0$.






                share|cite|improve this answer












                It may work. I assume $A$ is nonsingular. Suppose the true solution is $x_0$ and you are at $x = x_0 + y$ with $y$ small. Let $J$ be the Jacobian of $b$ at $x = x_0$.

                Then you get $w = A^{-1} b(x_0 + y) approx A^{-1}(b(x_0) + J y) = x_0 + A^{-1} J y$. If all eigenvalues of $A^{-1} J$ have absolute value less than $1$, you will converge to the solution $x_0$ provided you start close enough. If there is an eigenvalue with absolute value $> 1$, you will almost certainly not converge to $x_0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 16 at 13:30









                Robert Israel

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