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I am stuck in part B of this exercise (I am giving the whole exercise though). I would really appreciate it if someone could explain to me how to solve this type of integral. Thanks a lot!

Question: The usual way of deriving Adams-Bashforth multistep methods is to pass an interpolating polynomial through the previous $k$ values of the derivative, then integrate this from $t_{n+k-1}$ to $t_{n+k}$. An alternative approach, which leads to the Nystrom methods, is to start from the identity $$y(t_{n+k})=y(t_{n+k-2})+\int_{t_{n+k-2}}^{t_{n+k}}f(\tau,y(\tau))d\tau$$ which is satisfied by the exact solution. Then replace $f$ by the same interpolating polynomial (the one through the previous $k$ points), then integrate to get a formula for calculating $y_{n+k}$.

(A) With $k=2$, find the polynomial $P$ that interpolates $(t_n,y_n)$, $(t_{n+1},y_{n+1})$.

Answer: $$P(t)=y_n+(t-t_n)\frac{y_{n+1}-y_n}{h}$$

(B) Replace $f$ in the identity above with $P$ from part (a), and use this to obtain a numerical method of the form $$y_{n+2}-y_n=h\left(\beta_0 y_n'+\beta_1 y_{n+1}'+\beta_2 y_{n+2}'\right)$$ Explicitly calculate the coefficients $\beta_0$, $\beta_1$ and $\beta_2$. Then verify the order of the method is 2.

Answer: $$y_{n+2}-y_n=\int_{t_n}^{t_{n+2}}P(t)dt=\int_{t_n}^{t_{n+2}}\left[y_n+(t-t_n)\frac{y_{n+1}-y_n}{h}\right]dt$$

and I get lost there Apparently the result is:

$$=2hy_n+2h(y_{n+1}-y_n)=2hy_{n+1}$$

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Try this, remember $t_{n+2}=t_n+2h$ and everything can be treated as constants except for $t$. $$\begin{align}&\int^{t_{n+2}}_{t_n}\left[y_n+(t-t_n)\frac{y_{n+1}-y_n}{h}\right]\mathrm dt\\ &=y_n\int^{t_{n+2}}_{t_n}1\mathrm\,dt+\frac{y_{n+1}-y_n}{h}\int^{t_{n+2}}_{t_n}t-t_n\,\mathrm dt\\ &=2hy_n+\frac{y_{n+1}-y_n}{h}\left[\frac{1}{2}t^2-t_nt\right]^{t_{n+2}}_{t=t_n}\\ &=2hy_n+\frac{y_{n+1}-y_n}{h}\left[\frac{1}{2}t_{n+2}^2-t_nt_{n+2}-\frac{1}{2}t_n^2+t_n^2\right]\\ &=2hy_n+\frac{y_{n+1}-y_n}{h}\left[\frac{1}{2}\left(t_{n+2}+t_n\right)\left(t_{n+2}-t_n\right)+t_n\left(t_n-t_{n+2}\right)\right]\\ &=2hy_n+\frac{y_{n+1}-y_n}{h}\left[h\left(2t_{n}+2h\right)-2ht_n)\right]\\ &=2hy_n+\frac{y_{n+1}-y_n}{h}\left[2h^2\right]\\ &=2hy_n+2h\left(y_{n+1}-y_n\right)\\ &=2hy_{n+1}\end{align}$$

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  • $\begingroup$ Thanks a lot, it makes sense now... I was trying to integrate $t_n$ with respect to $t$. $\endgroup$ – s1047857 May 11 '15 at 8:35

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