2D Finite Difference Method

%% Finite Difference Method

syms x1 x2

x1_0 = 1;

x2_0 = 1;

tol = [0.0001;0.0001];

h = 0.1;

a = zeros(2,10);

a(1,1) = x1_0;

a(2,1) = x2_0;

f1 = @(x1,x2)(x1^2 +x2 -5);

f2 = @(x1,x2)(x2^2 -x1);

tic

for i = 1:100

K_11 = (f1(a(1,i)+h,a(2,i))- f1(a(1,i)-h,a(2,i)))/(2*h);

K_22 = (f2(a(1,i)+h,a(2,i))- f2(a(1,i)-h,a(2,i)))/(2*h);

K_12 = (f1(a(1,i),a(2,i)+h)- f1(a(1,i),a(2,i)-h))/(2*h);

K_21 = (f2(a(1,i),a(2,i)+h)- f2(a(1,i),a(2,i)-h))/(2*h);

J = [K_11 K_12; K_21 K_22];

a(:,i+1) = a(:,i) - inv(J)*[f1(a(1,i),a(2,i));f2(a(1,i),a(2,i))]; %%error line

if abs([(f1(a(1,i+1),a(2,i+1))); f2(a(1,i+1),a(2,i+1))]) < tol

fprintf('x1=%.8f & x2 = %.8f',a(1,i+1),a(2,i+1))

fprintf('no of itterations %d',i)

break

end

end

t = toc

I am trying to find roots for f1 and f2 with finite difference method. But I am getting this erro "Warning: Matrix is singular, close to singular or badly scaled. Results may be inaccurate. RCOND = NaN. > In NR (line 47)" the line 47 i have mentioned as %%error line in code.

same question i have done with 2D newton method and within 5 itteration i got the answer. (x1 = 1.90278368 & x2 = 1.37941426).

please guide me through this.

Thanks in advance.

Answer 1

[Torsten]

K_11 = (f1(a(1,i)+h,a(2,i))- f1(a(1,i)-h,a(2,i)))/(2*h);
K_22 = (f2(a(1,i),a(2,i)+h)- f2(a(1,i),a(2,i)-h))/(2*h);
K_12 = (f1(a(1,i),a(2,i)+h)- f1(a(1,i),a(2,i)-h))/(2*h);
K_21 = (f2(a(1,i)+h,a(2,i))- f2(a(1,i)-h,a(2,i)))/(2*h);

instead of

K_11 = (f1(a(1,i)+h,a(2,i))- f1(a(1,i)-h,a(2,i)))/(2*h);
K_22 = (f2(a(1,i)+h,a(2,i))- f2(a(1,i)-h,a(2,i)))/(2*h);
K_12 = (f1(a(1,i),a(2,i)+h)- f1(a(1,i),a(2,i)-h))/(2*h);
K_21 = (f2(a(1,i),a(2,i)+h)- f2(a(1,i),a(2,i)-h))/(2*h);   

[Dhruvilkumar Chavada]

Thanks a lot; now its showing same result as newtons method