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

Question

Ali Almakhmari 2023-2-12

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此问题的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1911085-newton-raphson-method-for-a-system-of-non-linear-equations

评论： Torsten 2023-2-12

采纳的回答： Alan Stevens

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.

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Answer 1

Alan Stevens 2023-2-12

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链接

此回答的直接链接

https://ww2.mathworks.cn/matlabcentral/answers/1911085-newton-raphson-method-for-a-system-of-non-linear-equations#answer_1169665

在 MATLAB Online 中打开

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

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Alan Stevens 2023-2-12

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 2023-2-12

在 MATLAB Online 中打开

@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