{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

# 216 in subsequent sections problems 1 problem 1 derive

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: e rst multistep formulas that we consider are explicit methods called Adams-Bashforth methods. These are derived by integrating the ODE y0 = f (t y) on the interval (tn;1 tn) to obtain Z or tn tn;1 y0dt = Z tn tn;1 y(tn) = y(tn;1) + Z (5.3.1) f (t y(t))dt tn tn;1 f (t y(t))dt: (5.3.2) Speci c numerical techniques are obtained by approximating f (t y(t)) in (5.3.2) by an interpolating polynomial and integrating the result. If, for example, f (t y(t)) were approximated by (the constant interpolating polynomial) f (t y(tn;1)) then (5.3.2) would produce Euler's method yn = yn;1 + hf (tn;1 yn;1): More generally, we'll interpolate f (t y(t)) by a k ; 1 st degree polynomial passing through tn;1, tn;2, : : : , tn;k . We'll use the Newton backward-di erence form of the interplating polynomial. For this application, we identify interpolation point tk in (5.2.16) with tn;1 , tk;1 with tn;2, : : : , and t1 with tn;k . Then, (5.2.16) becomes i;1 k ;1 X ri fn;1 Y (t ; tn;1;j ): Pk;1(t) = i i=0 i!h j =0 (5.3.3a) Using (5.2.11) with k replaced by k ; 1 and re-indexing the interpolation points as described above, we determine the error of this interpolation as Y f (k) ( y( )) k;1(t ; t Ek;1(t) = f (t y(t)) ; Pk;1(t) = n;1;j ) k! j =0 2 (tn;k tn;1): Let's expand both (5.3.3a) and (5.3.3b) to reveal their structure 2 n n Pk;1(t) = fn;1 + rfh ;1 (t ; tn;1) + r2!fh2;1 (t ; tn;1)(t ; tn;2 ) 13 (5.3.3b) k ;1 + : : : + r fn;1 1 (t ; tn;1 )(t ; tn;2) : : : (t ; tn;k+1) (k ; 1)!hk; and (k) Ek;1(t) = f ( k!y( )) (t ; tn;1)(t ; tn;2) : : : (t ; tn;k ): Using (5.2.14, 5.2.15), the rst few divided di erences are r0 fn;1 = fn;1 rfn;1 = fn;1 ; fn;2 r2 fn;1 = rfn;1 ; rfn;2 = fn;1 ; 2fn;2 + fn;3: The integration (5.3.2) can be simpli ed by the change of variables = t ; tn;1 h (5.3.4a) Since tn;1;j = tn;1 ; jh, we have t ; tn;1;j = ( + j )h and i;1 i;1 Y 1 Y(t ; t n;1;j ) = ( + j ) = ( + 1) : : : ( + i ; 1): hi j=0 j =0 Recall the combination symbol i = = ( ; 1) : : : ( ; i + 1) i!( ; i)! i! ! 0 = 1: (5.3.4b) This formula makes sense when is negative, i.e., ; i = ; (; ; 1) : : : (; ; i + 1) = (;1)i ( + 1) : : : ( + i ; 1) : i! i! Thus, i;1 1 Y(t ; t i; n;1;j ) = (;1) i i i!h j=0 : (5....
View Full Document

{[ snackBarMessage ]}

Ask a homework question - tutors are online