Newton Raphson loop for backward Euler

Dear all,

I would like to use Newton-Raphson method with backward Euler to meet a specific tolerance. How to change the loop below to get it?

% time step
h = (t_final - t_init)/n; % with n number of time steps

% vectors
t = [tinit zeros(1,n)]; % time
y = [yinit zeros(1,n)]; % solution

% Backward Euler loop
for i = 1:n
   t(i+1) = t(i) + h;
   y_temp = y(i) + h(f(t(i), y(i)));
   y(i+1) = y(i) + hf(t(i+1), y_temp);
end

I would do something like:

for i = 1:n

   error = 1;
   tolerance = 1e-6;

   t(i+1) = t(i) + h;
   y_temp = y(i) + h(f(t(i), y(i)));

   while error >= tolerance 
      y(i+1) = y(i) + hf(t(i+1), y_temp);
      error = abs(y(i+1) - y_temp) % (local) absolute error
      y_temp = y(i+1);
   end
end

Is this the correct approach? Could someone give an hint? Thanks in advance.

Answer 1

[Torsten]

You use a simple fixed-point iteration to get y(i+1), no Newton-Raphson.

Best wishes

Torsten.

[Clint_E]

Thanks a lot, so please how would you change it to get Newton-Raphson? Best wishes

[Torsten]

yold = y(i)+h*f(t(i),y(i));

while error >= tolerance

ynew = yold-(yold-(y(i)+h*f(t(i+1),yold)))/(1-h*df(t(i+1),yold));
error = abs(ynew-yold);
yold=ynew;

end

y(i+1) = ynew;

df is the derivative of f with respect to y.

Best wishes

Torsten.

[Clint_E]

Thanks a lot! Best wishes