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1. The nonlinear system fi(x1, x2) = x - xż + 2x2 = 0, $2(x1, x2) = 2x1 + xź – 6= 0 has two solutions, (0.625204094, 2.179355Continuation Algorithm To approximate the solution of the nonlinear system F(x) = 0 given an initial approxima- tion x: INPUTGiven this pseudocode and problem (as an example), code the continuation algorithm in MATLAB.

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Answer #1

%%%%%%%%% With initial condition x1=0, x2=0 %%%%%%

clc;
clear all;

N=2;

f1=@(x1,x2) x1^2-x2^2+2*x2;
f2=@(x1,x2) 2*x1+x2^2-6;
J=@(x1,x2)[2*x1 -2*x2+2;2 2*x2];
%intial conditions
x1(1)=0;
x2(1)=0;
F=[f1(x1,x2);f2(x1,x2)];
%step 1
h=1/N;
b=-h*F;
% Step 2
for i=1:2
A=J(x1(i),x2(i));
K1=inv(A)*b;
%step 3
A1=J(x1(i)+K1(1)/2,x2(i)+K1(2)/2);
K2=inv(A1)*b;
%step 4
A2=J(x1(i)+K2(1)/2,x2(i)+K2(2)/2);
K3=inv(A2)*b;
%step 5
A3=J(x1(i)+K3(1),x2(i)+K3(2));
K4=inv(A3)*b;
x1(i+1)=x1(i)+(K1(1)+2*K2(1)+2*K3(1)+K4(1))/6;
x2(i+1)=x2(i)+(K1(2)+2*K2(2)+2*K3(2)+K4(2))/6;
end
x1=x1(end)
x2=x2(end)
%%%%%%%%% Answer %%%%%%%%


x1 =

2.1195


x2 =

-1.3723

>>

%%%%%%%% With initial condition x1=1, x2=2 %%%%%%%%%%

clc;
clear all;

N=2;

f1=@(x1,x2) x1^2-x2^2+2*x2;
f2=@(x1,x2) 2*x1+x2^2-6;
J=@(x1,x2)[2*x1 -2*x2+2;2 2*x2];
%intial conditions
x1(1)=1;
x2(1)=1;
F=[f1(x1,x2);f2(x1,x2)];
%step 1
h=1/N;
b=-h*F;
% Step 2
for i=1:2
A=J(x1(i),x2(i));
K1=inv(A)*b;
%step 3
A1=J(x1(i)+K1(1)/2,x2(i)+K1(2)/2);
K2=inv(A1)*b;
%step 4
A2=J(x1(i)+K2(1)/2,x2(i)+K2(2)/2);
K3=inv(A2)*b;
%step 5
A3=J(x1(i)+K3(1),x2(i)+K3(2));
K4=inv(A3)*b;
x1(i+1)=x1(i)+(K1(1)+2*K2(1)+2*K3(1)+K4(1))/6;
x2(i+1)=x2(i)+(K1(2)+2*K2(2)+2*K3(2)+K4(2))/6;
end
x1=x1(end)
x2=x2(end)


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


x1 =

0.6213


x2 =

2.1899

>>

%%%%%%%%%%%% With initial condition x1=3, x2=-2 %%%%%%%%%

clc;
clear all;

N=2;

f1=@(x1,x2) x1^2-x2^2+2*x2;
f2=@(x1,x2) 2*x1+x2^2-6;
J=@(x1,x2)[2*x1 -2*x2+2;2 2*x2];
%intial conditions
x1(1)=3;
x2(1)=-2;
F=[f1(x1,x2);f2(x1,x2)];
%step 1
h=1/N;
b=-h*F;
% Step 2
for i=1:2
A=J(x1(i),x2(i));
K1=inv(A)*b;
%step 3
A1=J(x1(i)+K1(1)/2,x2(i)+K1(2)/2);
K2=inv(A1)*b;
%step 4
A2=J(x1(i)+K2(1)/2,x2(i)+K2(2)/2);
K3=inv(A2)*b;
%step 5
A3=J(x1(i)+K3(1),x2(i)+K3(2));
K4=inv(A3)*b;
x1(i+1)=x1(i)+(K1(1)+2*K2(1)+2*K3(1)+K4(1))/6;
x2(i+1)=x2(i)+(K1(2)+2*K2(2)+2*K3(2)+K4(2))/6;
end
x1=x1(end)
x2=x2(end)


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


x1 =

2.1095


x2 =

-1.3345

>>

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