I have the following initial value problem on $t \in [0,T]$: $$y'=y^2(1-\epsilon y)$$
I wish to use the Trapezoidal method to solve this I.V.P., and I cannot use simply use the predictor-corrector method, i.e. improved Euler's method. The trapezoidal method is defined by: $$y_{n+1}=y_n+\frac{h}{2}(f_{n+1}+f_n)$$ Plugging in the value from the O.D.E., we have after some simplification: $$y_{n+1}=y_n+\frac{h}{2}(y_{n+1}^2+y_n^2-\epsilon y_{n+1}^3-\epsilon y_n^3)$$
This is an implicit method, so I must use a root finding method at each iteration to find the $y_{n+1}$ on the right side of the equation. I have read online that Newton's method is usually used for such purposes, but I cannot figure out how to implement it.
$$y_{n+1}=y_n-\frac{f(y_n)}{f'(y_n)}$$
For example, what values would $f(y_n)$ and $f'(y_n)$ take in this context? Thank you in advance!
Best Answer
Your whole equation is $f$. More precisely, $y_{n+1}$ is the solution of $$ 0=f(y)=y-y_n-\frac h2(y^2+y_n^2-ϵ(y^3+y_n^3)) $$ closest to $y_n$.
The derivative is then $$ f'(y)=1-\frac h2(2y-3ϵy^2) $$ so that the Newton iteration is $$ y_+=y-\frac{f(y)}{f'(y)}=\frac{y_n-\frac h2(y^2-y_n^2−ϵ(2y^3-y_n^3))}{1-\frac h2(2y-3ϵy^2)} $$