# System of nonlinear equations F(X) = 0

## Newton's method

Problem: Find the solution of the nonlinear system

x_{2} = x_{1}^{3} + 1

x_{1}^{2} + x_{2}^{2} = 4

First of all, transform the system to the form F(X) = 0,

where
X = [ x_{1} , x_{2} ]^{T},
F = [ f_{1}(X) , f_{2}(X) ]^{T} :

f_{1}(X) = x_{1}^{2} + x_{2}^{2} - 4 ,

f_{2}(X) = x_{2} - x_{1}^{3} - 1 .

Using Matlab, define F as

>> syms x1 x2 % define symbolic variables x1, x2 >> XV = [ x1 ; x2] % vector of the variables x1, x2 >> f1 = x1^2 + x2^2 -4 % function f1 of x1, x2 >> f2 = x2-x1^3-1 % function f2 of x1, x2 >> F = [ f1 ; f2 ] % vector function F of x1, x2

Jacobi matrix of F:

>> J = jacobian(F, XV)

Choose a first approximation X:

>> X = [1; 2]

Repeat following 4 commands:

>> JX = subs( J, {x1, x2}, {X(1), X(2)}) % substitute a value of X to J >> FX = subs( F, {x1, x2}, {X(1), X(2)}) % substitute a value of X to F >> D = - JX \ FX % solve linear system JX * D = - FX >> X = X + D % upgrade the approximation X

## Fixed point iterations

Solving the same problem as above using fixed point iterations.

Using Matlab, define F as above:

>> syms x1 x2 % define symbolic variables x1, x2 >> X = [ x1 ; x2] % vector of the variables x1, x2 >> f1 = x1^2 + x2^2 -4 % function f1 of x1, x2 >> f2 = x2-x1^3-1 % function f2 of x1, x2 >> F = [ f1 ; f2 ] % vector function F of x1, x2

Fixed point iterations:

>> maxit = 20; % number of iterations >> X = [1; 2]; % a first approximation >> alfa = 0.1; % a coefficient for improving convergency - a guess >> G = [x1; x2] - alfa*F; % transformation of F(x) = 0 to x = G(x) >> for k=1:maxit >> X = subs(G, {x1, x2}, {X(1), X(2)}); % x_{k} = G(x_{k-1}) >> % displaying of the last 3 iterations: >> if (k > (maxit-3)) >> if(k == (maxit-2)) >> disp('the last 3 approximations of X:') >> end >> disp(X) >> end >> end >> % test of the result by computing F(X) - should be close to zero >> FX = subs(F, {x1, x2}, {X(1), X(2)})

Experiment with different values of maxit (number of iterations), X (the first approximation) and alfa (the coefficient influencing convergency).

Solve a similar problem with second equation changed to

x_{2} = sin(x_{1}) + 1

so that

f2 = x2 - sin(x1) - 1 ;

(now the better choice of the parameter alfa would be 0.25)