I have the following non-linear system to solve with Newton's method in matlab:
x²+y²=2.12
y²-x²y=0.04
What is the linear equation system to be solved? Should I calculate the Jacobian matrix? I should use the starting variables x=1, y=1 and solve it with a matlab program with an error less than 10^-6. .Can you tell me how to do it?
Update
Here is my solution that is the same as matlab's solver. It's just like Newton for scalar and in vector form:
>>f=inline('x(1)^2+x(2)^2-2.12; x(2)^2-x(1)^2*x(2)-0.04]');
>>x=fsolve(f,x)
equation solved
x=
1.006852730995833
1.051783055209322
>>while dxnorm>0.4e-6 & iter<10
f=[x(1)^2+x(2)^2-2.12;x(1)^1-x(1)^2*x(2)-0.4];
J=[2*x(1) 2*x(2);-2*x(1)*x(2) 2*x(2)*x(1)^2];
dx=-J\f
x=x+dx
dxnorm=norm(dx,inf), iter=iter+1;
end
>>x, iter
x=1.006852730995833
1.051783055209322
Best Answer
Unfortunately, your update doesn't convince me you did the correct job, based on your script, your
x
variable is already the answer and residing in the workspace, the second time when your own scripts are being executed, only one iteration has been made, not the whole Newton's method.If you are allowed to use
Symbolic Toolbox
to compute the Jacobian, there is a more universal way of doing instead of just carrying out a problem-oriented script. Here is a modified version to match your notation of an old implementation of mine for Newton's method, and this could be easily vectorized for a multi-dimensional nonlinear equation system usingvarargin
input, and do a string size check on the inline function you passed to the following function.This function accepts input
f
as an inline function,p0
is the initial guess,tol
is the tolerance andMaxIter
is the maximum iteration allowed, if the vectorized version of your input is:Then
would return:
As you can see the last output is different from yours, because basically what you had was
fsolve
output, howeverfsolve
usetrust-region-dogleg
algorithm by default, not Newton's method, if your output coincides withfsolve
's result, then your implementation should have a problem.