Newton's Method for nonlinear system vector operation.

when i was doing newton's method for nonlinear system, when I entered following code it tells me that it could not do subtraction between two vectors with different dimension. The thing is F is a 2x1 vector, and J is jacobian matrix of F which is 2x2. so I dont know what is going on with my code. the following is the code.

    a = newton(@F, @J, [1,1])
=========================================  
  function y = newton(f,j,x) 

step = 1;
finalstep = 10;
y = zeros (2,1);

while step < finalstep
    d = j(x)\(-f(x));

      y = x + d;
      disp(y);
      x = y;
  end
  end
==========================================
  function f = F(x)
  x1 = x(1);
  x2 = x(2);
  f = zeros(2,1);
  f(1) = x1^2-x2^2+2*x2; % f1(x1,x2)
  f(2) = 2*x1+x2^2-6; % f2(x1,x2);
  end
==========================================
  function j = J(x)
  x1 = x(1);
  x2 = x(2);
  j = zeros(2,2);
  j(1,1) = 2*x1; % df1x1
  j(1,2) = -2*x2+2; % df1x2
  j(2,1) = 2; % df2x1
  j(2,2) = 2*x2; % df2x2;
  end

Answer 1

[Geoff Hayes]

Lechen - if I run through your code, I observe the following error

 Error using  + 
 Matrix dimensions must agree.

 Error in newton.m
        y = x + d;

because x is a 1x2 matrix and d is a 2x1 matrix. The simplest fix for this is to just change the input x from a 1x2 matrix to one that is 2x1

 a = newton(@F, @J, [1;1])

That will fix the error message. Now look closer at the newton function

 function y = newton(f,j,x) 
      step = 1;
      finalstep = 10;
      y = zeros (2,1);

      while step < finalstep
          d = j(x)\(-f(x));
          y = x + d;
          disp(y);
          x = y;
      end
 end

Note how the code does not increment the step local variable, so the code will become stuck in an infinite loop. Add a line to increment this variable. Also, consider adding it a line of code that compares x and y - if the difference between the two vectors (for each element) is less than some tolerance, then assume that no better solution will be found and so break out of the loop. The above then becomes something like

 function y = newton(f,j,x) 
      step = 1;
      finalstep = 10;
      y   = zeros (2,1);
      tol = 1e-5; 
      while step < finalstep
          d = j(x)\(-f(x));
          y = x + d;
          disp(y);

          % increment step
          step = step + 1;

          % compare each element of x and y
          exitEarly = true;
          for k=1:length(x)
              if abs(x(k)-y(k))>tol
                  % pair k exceeds tolerance so do not exit
                  % early
                  exitEarly=false;
              end
          end

          if exitEarly
              break;
          end          

          x = y;

      end
 end

Try the above and see what happens!

[Lechen Yuan]

for the love of mother of god, finally........ i was trying to fix the x into a column vector the whole time! and never thought about fixing it outside of the script. Thank you so much!!!!!!!!!!

Answer 2

[Andrei Bobrov]

function x = newtonf(x,finalstep) 
    function out = F(x)
        out = [x(1).^2-x(2).^2+2*x(2);
               2*x(1)+x(2).^2-6];
    end

      function out = J(x)
          out = [2*x(1),-2*x(2)+2;
                 2,      2*x(2)  ];
      end

      step = 1;
      y = [];
      while step < finalstep
          d = -J(x)\F(x);
          x = x + d;
          y = [y, x];
          step = step + 1;
      end
      disp(y);
  end