# Newton-Raphson Method for Non-linear System of 3 variables in Matlab

1. Jun 17, 2014

### wel

I am trying to solve 3 non-linear system of 3 variables using the newton-raphson method in matlab. Here are the 3 non-linear equations:

$$c[\alpha I+ k_f+k_d+k_ns+k_p(1-q)]-I \alpha =0$$
$$s[\lambda_b c P_C +\lambda_r (1-q)]- \lambda_b c P_C =0$$
$$q[\gamma +c k_p \frac{P_C}{P_Q}]- c k_p \frac{P_C}{P_Q}=0$$

I need to find the values of c,s, and q using the newton-raphson method.

=>
This is my matlab code :

Code (Text):
format long
clear;

%values of parameters
I=1200;
k_f= 6.7*10.^7;
k_d= 6.03*10.^8;
k_n=2.92*10.^9;
k_p=4.94*10.^9;
lambda_b= 0.0087;
lambda_r =835;
gamma =2.74;
alpha =1.14437*10.^-3;
P_C= 3 * 10.^(11);
P_Q= 2.87 * 10.^(10);
tol = 10.^-4;  %tol is a converge tolerance
%initial guess or values
c=1;
s=0.015;
q=0.98;
iter= 0; %iterations
xnew =[c;s;q];
xold = zeros(size(xnew));
while norm(xnew - xold) > tol
iter= iter + 1;
xold = xnew;      % update c, s, and q
c = xold(1);
s = xold(2);
q = xold(3);

%Defining the functions for c,s and q.
f = c * (alpha*I + k_f + k_d + k_n * s + k_p*(1-q))-I *alpha;
g = s * (lambda_b * c* P_C + lambda_r *(1-q))- lambda_b* c * P_C;
h = q * ( gamma + c * k_p *(P_C / P_Q))- (c * k_p * (P_C / P_Q));

%Partial derivatives in terms of c,s and q.
dfdc = alpha*I + k_f + k_d + k_n * s + k_p*(1-q);
dfds = k_n *c ;
dfdq = - k_p *c;

dgdc = lambda_b * P_C *(s-1);
dgds = lambda_b * c* P_C + lambda_r *(1-q);
dgdq = - lambda_r * s;

dhdc = k_p *(P_C / P_Q)*(q-1);
dhds = 0;
dhdq = gamma + c * k_p *(P_C / P_Q);

%Jacobian matrix
J = [dfdc dfds dfdq; dgdc dgds dgdq; dhdc dhds dhdq];

% Applying the Newton-Raphson method
xnew = xold - J\[f;g;h];
disp(sprintf('iter=%6.15f,  c=%6.15f,  s=%6.15f, q=%6.15f', iter,xnew));
end

can someone please check my code, there are no errors but did i got accurate values of c,s, and q?. Thanks in advance.

Last edited: Jun 17, 2014
2. Jun 17, 2014

### phyzguy

I'm not a Matlab expert, and I didn't check your code in detail, but clearly your definition of the Jacobian is wrong.

You have: J = [dfdc dfds dfds; dgds dgds dgds; dhds dhds dhds];

You want: J = [dfdc dfds dfdq; dgdc dgds dgdq; dhdc dhds dhdq];