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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!

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1 Answer 1

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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 $$ y_+=y-\frac{f(y)}{f'(y)}=\frac{y_n-\frac h2(y^2-y_n^2−ϵ(2y^3-y_n^3))}{1-\frac h2(2y-3ϵy^2)} $$

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  • $\begingroup$ I don't quite understand. So, for each iteration, $y_{n+1}=y-y_n-\frac{h}{2}(y^2+y_n^2-\epsilon(y^3+y_n^3))$ ? Also, what would $f'(y_n)$ be? $\endgroup$
    – user208614
    Nov 8, 2015 at 23:52
  • $\begingroup$ No, why? That would imply $y_{n+1}=0$. $\endgroup$
    – Lutz Lehmann
    Nov 8, 2015 at 23:55
  • $\begingroup$ Sorry, I'm still unsure of how to apply Newton's method to this concept. I know I'm supposed to use it to find an expression for $y_{n+1}$ on the right side. $\endgroup$
    – user208614
    Nov 8, 2015 at 23:58
  • $\begingroup$ Thank you, that helps a lot! I'm just confused about how the $\epsilon(2y^2-y_n^2)$ term on the numerator. How did those powers of $3$ cancel out? $\endgroup$
    – user208614
    Nov 9, 2015 at 0:34
  • $\begingroup$ Sorry, they don't. $\endgroup$
    – Lutz Lehmann
    Nov 9, 2015 at 0:49

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