Newton Raphson method for a system of non-linear equations?

If I have a non-linear system of two equations that are each in terms of two variables x and y:

eqn1(x,y) = 0

eqn2(x,y) = 0

How will I go about solving those two equations using Newton-Raphson in MATLAB? I know how to solve this system using the MATLAB's Solve function but not Newton-Raphson.

Answer 1

[Alan Stevens]

Here's a simple example for you to follow:

% Functions
f = @(x,y) x^2 + y - 5.94;  % i.e. the equation is x^2+y=5.94
g = @(x,y) x + y^2 - 3.41;  % similarly this represents equation x+y^2=3.41
F = @(x,y)[f(x,y); g(x,y)];
% Jacobian
% J = [df/dx, df/dy;
%      dg/dx, dg/dy];
J = @(x,y) [2*x 1; 1 2*y];

% Initial guess
xy = [1; 1];

% Tolerance
tol = 10^-6;

err = 1;
while err>tol
      
      xynew = xy - J(xy(1),xy(2))\F(xy(1),xy(2));
      err = abs(xynew-xy);
      xy = xynew;
      
end
disp(xy)

    2.2000
    1.1000

[Ali Almakhmari]

Thank you so much! One more question, if I have the jacobian matrix defined symbolically and stored in a variable called "j", how would do I go about assigning it to J? I tried J = @(x,y) j, but J was not evaluted with my variable j.

[Alan Stevens]

Since Newton-Raphson is essentially numerical it seems pointless to use symbolics at all!

However, you might try J = @(x,y) j(x,y)

I don't have the symbolic toolbox, so can't actually check!

[Torsten]

@Ali Almakhmari

% Functions
syms x y
f = x^2 + y - 5.94;  % i.e. the equation is x^2+y=5.94
g = x + y^2 - 3.41;  % similarly this represents equation x+y^2=3.41
F = [f; g];
% Jacobian
% J = [df/dx, df/dy;
%      dg/dx, dg/dy];
J =  jacobian(F);

F = matlabFunction(F)

F = function_handle with value:
    @(x,y)[y+x.^2-2.97e+2./5.0e+1;x+y.^2-3.41e+2./1.0e+2]

J = matlabFunction(J)

J = function_handle with value:
    @(x,y)reshape([x.*2.0,1.0,1.0,y.*2.0],[2,2])

% Initial guess
xy = [1; 1];

% Tolerance
tol = 10^-6;

err = 1;
while err>tol
      
      xynew = xy - J(xy(1),xy(2))\F(xy(1),xy(2));
      err = abs(xynew-xy);
      xy = xynew;
      
end
disp(xy)

    2.2000
    1.1000

disp(F(xy(1),xy(2)))

   1.0e-15 *

    0.8882
         0