Question

Use Matlab to solve the system x²+xy³=9 , 3x²y-y³ =4 using Newton's method for nonlinear system. use each of initial guesses (x₀,y₀)=(1.2,2.5), (-2,2.5), (-1.2,-2.5), (2,-2.5) observe which root to which the method converges, the number of iterates required and the speed of convergence.

Write the system in the form f(u) = 0, and report for each case the number of iterations needed for ||f(u)||²≤ 10^-12−.

Answer 1

clc;
clear all;

%intial condition
x=1.2;
y=2.5;

%tolerence
tol=1e-12;
err=0.5;
%%%Initial points save in x1and x2 for print purpose
x1(1)=x;
x2(1)=y;
n=2;
while(err>tol)
%functio F
f1= x.^2+x.*y.^3-9;
f2=3*x.^2.*y-y.^3-4 ;
  
  
%%% partial derivatives of F (linearization)
f11=2*x+y.^3;
f12=3*y.^2.*x;

f21=6*x.*y;
f22=3*x.^2-3*y.^2;

J=[f11 f12 ; f21 f22 ]; %jacobian matrix
J1=inv(J);
Z=[x ;y]-J1*[f1; f2]; % newton formula
err=norm([f1; f2],2);
x=Z(1);
y=Z(2);

% %%% Every iteration values save in x1 and x2 for print
x1(n)=x;
x2(n)=y;
n=n+1;

end
n
disp('____________________________________________________________________')
disp('iteration x1 x2 ')
disp('____________________________________________________________________')
for i=1:n-1
  
fprintf('%d\t %15f \t %15f \n',i,x1(i),x2(i))
end
disp( 'Intersection point')
[x,y]
disp('Number of iterations with initial condition (1.2, 2.5)')
n-1

clc;
clear all;

%intial condition
x=-2;
y=2.5;

%tolerence
tol=1e-12;
err=0.5;
%%%Initial points save in x1and x2 for print purpose
x1(1)=x;
x2(1)=y;
n=2;
while(err>tol)
%functio F
f1= x.^2+x.*y.^3-9;
f2=3*x.^2.*y-y.^3-4 ;
  
  
%%% partial derivatives of F (linearization)
f11=2*x+y.^3;
f12=3*y.^2.*x;

f21=6*x.*y;
f22=3*x.^2-3*y.^2;

J=[f11 f12 ; f21 f22 ]; %jacobian matrix
J1=inv(J);
Z=[x ;y]-J1*[f1; f2]; % newton formula
err=norm([f1; f2],2);
x=Z(1);
y=Z(2);

% %%% Every iteration values save in x1 and x2 for print
x1(n)=x;
x2(n)=y;
n=n+1;

end

disp('____________________________________________________________________')
disp('iteration x y ')
disp('____________________________________________________________________')
for i=1:n-1
  
fprintf('%d\t %15f \t %15f \n',i,x1(i),x2(i))
end
disp( 'Intersection point')
[x,y]
disp('Number of iterations with initial condition (-2, 2.5)')
n-1

%%%%%%%%%%%%%%%%%%%% Answer

____________________________________________________________________
iteration x y   
____________________________________________________________________
1   -2.000000    2.500000
2   -1.473397    1.696580
3   -1.179337    0.639062
4   -3.195436    -1.913207
5   -1.742641    -1.795509
6   -0.971947    -1.974885
7   -0.895735    -2.089564
8   -0.901263    -2.086579
9   -0.901266    -2.086588
10   -0.901266    -2.086588
11   -0.901266    -2.086588
Intersection point

ans =

-0.9013 -2.0866

Number of iterations with initial condition (-2, 2.5)

ans =

11

>>

clc;
clear all;

%intial condition
x=-1.2;
y=-2.5;

%tolerence
tol=1e-12;
err=0.5;
%%%Initial points save in x1and x2 for print purpose
x1(1)=x;
x2(1)=y;
n=2;
while(err>tol)
%functio F
f1= x.^2+x.*y.^3-9;
f2=3*x.^2.*y-y.^3-4 ;
  
  
%%% partial derivatives of F (linearization)
f11=2*x+y.^3;
f12=3*y.^2.*x;

f21=6*x.*y;
f22=3*x.^2-3*y.^2;

J=[f11 f12 ; f21 f22 ]; %jacobian matrix
J1=inv(J);
Z=[x ;y]-J1*[f1; f2]; % newton formula
err=norm([f1; f2],2);
x=Z(1);
y=Z(2);

% %%% Every iteration values save in x1 and x2 for print
x1(n)=x;
x2(n)=y;
n=n+1;

end

disp('____________________________________________________________________')
disp('iteration x y ')
disp('____________________________________________________________________')
for i=1:n-1
  
fprintf('%d\t %15f \t %15f \n',i,x1(i),x2(i))
end
disp( 'Intersection point')
[x,y]
disp('Number of iterations with initial condition (-1.2, -2.5)')
n-1

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%% Answer

____________________________________________________________________
iteration x y   
____________________________________________________________________
1   -1.200000    -2.500000
2   -0.985132    -2.174800
3   -0.908497    -2.091489
4   -0.901311    -2.086601
5   -0.901266    -2.086588
6   -0.901266    -2.086588
7   -0.901266    -2.086588
Intersection point

ans =

-0.9013 -2.0866

Number of iterations with initial condition (-1.2, -2.5)

ans =

7

>>

clc;
clear all;

%intial condition
x=2;
y=-2.5;

%tolerence
tol=1e-12;
err=0.5;
%%%Initial points save in x1and x2 for print purpose
x1(1)=x;
x2(1)=y;
n=2;
while(err>tol)
%functio F
f1= x.^2+x.*y.^3-9;
f2=3*x.^2.*y-y.^3-4 ;
  
  
%%% partial derivatives of F (linearization)
f11=2*x+y.^3;
f12=3*y.^2.*x;

f21=6*x.*y;
f22=3*x.^2-3*y.^2;

J=[f11 f12 ; f21 f22 ]; %jacobian matrix
J1=inv(J);
Z=[x ;y]-J1*[f1; f2]; % newton formula
err=norm([f1; f2],2);
x=Z(1);
y=Z(2);

% %%% Every iteration values save in x1 and x2 for print
x1(n)=x;
x2(n)=y;
n=n+1;

end

disp('____________________________________________________________________')
disp('iteration x y ')
disp('____________________________________________________________________')
for i=1:n-1
  
fprintf('%d\t %15f \t %15f \n',i,x1(i),x2(i))
end
disp( 'Intersection point')
[x,y]
disp('Number of iterations with initial condition (2, -2.5)')
n-1

%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%

____________________________________________________________________
iteration x y   
____________________________________________________________________
1   2.000000    -2.500000
2   1.224118    -1.773857
3   0.352397    -0.769761
4   -22.175315    22.599350
5   -14.832213    17.565361
6   -10.202207    13.543333
7   -7.190606    10.367504
8   -5.168467    7.889394
9   -3.772849    5.973114
10   -2.789209    4.495752
11   -2.085083    3.345040
12   -1.572499    2.407123
13   -1.186599    1.518463
14   -0.939118    0.121361
15   -5.272400    0.397125
16   -3.491379    0.316557
17   -3.029761    0.192941
18   -3.001248    0.148966
19   -3.001625    0.148108
20   -3.001625    0.148108
21   -3.001625    0.148108
Intersection point

ans =

-3.0016 0.1481

Number of iterations with initial condition (2, -2.5)

ans =

21

>>