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# Higher order derivatives for N -body simulations

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Derivatives of Newtonian gravity.

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### Higher order derivatives for N -body simulations

1. 1. Higher order derivatives for N-body simulations Keigo Nitadori July 3, 2014 1 Derivatives of power functions Consider y = xn , (1) with time dependent x and y, and constant n. From lny =n lnx, (2) ˙y y =n ˙x x , (3) we have 0 =n ˙xy − x ˙y, 0 =n ¨xy + (n − 1) ˙x ˙y − x ¨y, 0 =n ... xy + (2n − 1) ¨x ˙y + (n − 2) ˙x ¨y − x ... y , 0 =n .... x y + (3n − 1) ... x ˙y + (3n − 3) ¨x ¨y + (n − 3) ˙x ... y − x .... y . (4) Here, ˙[ ] = d dt [ ] and we also note [ ](n) = dn dtn [ ]. From these, higher order derivatives of y are recursively available in ˙y =[n ˙xy]/x, ¨y =[n ¨xy + (n − 1) ˙x ˙y]/x ... y =[n ... xy + (2n − 1) ¨x ˙y + (n − 2) ˙x ¨y]/x, .... y =[n .... x y + (3n − 1) ... x ˙y + (3n − 3) ¨x ¨y + (n − 3) ˙x ... y ]/x, (5) and so on. A general form for k ≥ 1 is y(k) = 1 x k−1 i=0 Ck,ix(k−i) y(i) , (6) 1
2. 2. with coeﬃcients Ck,0 =n Ck,k = − 1 Ck,i =Ck−1,i−1 + Ck−1,i = k − 1 i n − k − 1 i − 1 (1 ≤ i ≤ k − 1). (7) 2 Derivatives of gravitational force Now we focus on the time derivatives of the gravitational force, f = m r r 3 . (8) Let s = (r · r), and q = s−3/2 , (9) then f =mqr, ˙f =mq ˙r + ˙q q r , ¨f =mq ¨r + 2 ˙q q ˙r + ¨q q r , ... f =mq ... r + 3 ˙q q ¨r + 3 ¨q q ˙r + ... q q r , (10) and f (k) =m k i=0 k i q(i) r(k−i) =mq k i=0 k i q(i) q r(k−i) . (11) For the derivatives of q, by using n = −3 2, we have q =s−3/2 , ˙q q = − 3 s ˙s 2 , ¨q q = − 3 s ¨s 2 + 5 3 ˙s 2 ˙q q ... q q = − 3 s ... s 2 + 8 3 ¨s 2 ˙q q + 7 3 ˙s 2 ¨q q . (12) The coeﬃcients multiplied by 6 are: 2
3. 3. @ @ @k i 0 1 2 3 4 5 1 3 (2) 2 3 5 (2) 3 3 8 7 (2) 4 3 11 15 9 (2) 5 3 14 26 24 11 (2) Derivatives of s have simple form in, s =(r · r), ˙s =2(r · ˙r), ¨s =2(r · ¨r) + 2( ˙r · ˙r), ... s =2(r · ... r ) + 6( ˙r · ¨r), s(4) =2(r · r(4) ) + 8( ˙r · ... r ) + 6( ¨r · ¨r), s(5) =2(r · r(5) ) + 10( ˙r · r(4) ) + 20( ¨r · ... r ), (13) and a general form is s(k) = k i=0 k i (r(i) · r(k−i) ) =    2 (k−1)/2 i=0 k i (r(i) · r(k−i) ) (k is odd)   2 (k−2)/2 i=0 k i (r(i) · r(k−i) )   + k k/2 (r(k/2) · r(k/2) ) (k is even) . (14) 3 Another approach Le Guyader (1993) took a slightly diﬀerent approach. For the derivatives of r = r , r2 = (r · r), (15) r ˙r = (r · ˙r), (16) and after (k − 1) times diﬀerentiations, k−1 i=0 k − 1 i r(i) r(k−i) = k−1 i=0 k − 1 i (r(i) · r(k−i) ). (17) Finally, we have r(k) = 1 r   (r · r(k) ) + k−1 i=1 k − 1 i (r(i) · r(k−i) ) − r(i) r(k−i)   , (18) 3
4. 4. for k ≥ 2. For the derivatives of q = r−3, from r ˙q = −3˙rq, (19) after diﬀerentiating (k − 1) times, k−1 i=0 k − 1 i r(i) q(k−i) = −3 k−1 i=0 k − 1 i r(k−i) q(i) . (20) Thus, q(k) = − 1 r   3r(k) q + k−1 i=1 k − 1 i 3r(k−i) q(i) + r(i) q(k−i)   , (21) for k ≥ 2. References Le Guyader, C. 1993, A&A, 272, 687. http://adsabs.harvard.edu/abs/1993A&A... 272..687L 4