6.15

Determine the roots of the following simultaneous nonlinear equations using 

(a) fixed-point iteration and (b) the newton-Raphson method.

 y = -x^2 + x + 0.75

 y + 5xy =x^2

Employ initial guesses of x = y = 1.2 and discuss the results.

x = -1:-.25;

y = -1:5;

fx = @(x,y) y + x .^2  - .75; %solved for y

fy = @(x,y) (-5 .* x .* y) + x .^2 ;%solved for y

%fy = @(x,y) (-5 .* y + 1) .* x ;%solved for y

%fy = @(x,y) -x .^2  + x + .75; %solved for y

%fx = @(x,y) sqrt((5 .* x .* y) + y) ;%solved for y

%fx = @(x,y) sqrt(-y + x  - .75); %solved for y

%fy = @(x,y) (x .^2)/ (1 + 5*x) ;%solved for y

hold on

plot (fy(x,y))

plot (fx(x,y))

% newton-raphson for systems of non-linear equations

u = @(x,y) - x .^2 + x + .75 - y; % u derivitation by parts

v = @(x,y) y + 5 .* x .* y - x .^2;%v derivitation by parts

dudx =@(x,y) -2 * x + 1;

dudy = @(x,y) -1;

dvdx = @(x,y) 5*y - 2*x;

dvdy = @(x,y) 1+ 5*x;

%y1 = fy(x,y); first iteration of y

%x1 = fx(x,y);first iteration of x

%y2 = fy(x1,y1); second iteration of y

%x2 = fx(x1,y1);second iteration of x

x = 1.2;

y = .2;

for ii = 1:3

   

    ny = fy(x,y);

    nx = fx(x,y);

    

   y = ny;

   x = nx;

   plot(x,y,'*r') 

end

%guess

x = 4; y = 0;

denom = ((dudx(x,y) * dvdy(x,y)) - (dudy(x,y) * dvdx(x,y)));

for ii = 0:15

denominator = denom + ( denom == 0) ;

x_1 = x - (((u(x,y) * dvdy(x,y)) - (v(x,y) * dudy(x,y))) / denominator);

y_1 = y - (((v(x,y) * dudx(x,y)) - (u(x,y) * dvdx(x,y))) / denominator);

x = x_1;

y = y_1;

plot( x_1 , y_1, '*g')

end

hold off

When running the fixed point at the initial guesses specified, it diverges very quickly, so we chose to start the y at .2 instead because it is closer. Nevertheless, even after bringing the initial guess closer, it diverges after 4 iterations. The newton Raphson method was also rather inaccurate at first, but x=4, y=0 seemed to get quite close, though still not quite there.