ff NASA-CR-194X3X / f J Stability by Linear Processes //C/-_/__/_// Problem: To find a quick way to determine if the origin 0 in the _--_ complex plane C within the image of Q under the multilinear function p(j_,q) = f(q) + g(q) j for fixed _. Notation: Parameter Space Rn C /_I_6_ i , AI,A2, ... in Q C R" A1r,_ I , • ,, /_ /_ are preimages of points in C AI', _' • . • under p(j_,q) . Il edge and 11',12 ', ...,in' Iz, ...i. line segments in C line segments where which are images of line 11 ± Iz ... ± I. segments It,12,...,l . respectively from Q. Without loss of generality, assume that the point (q1, on', ..., qn) and the point (ql+, q{, ..., q.-) are the endpoints of the edge iI and the point A l e iI would be (ql, on', --., qn) then the point (ql, q{, ..., qn) and the point (ql, on+, on" -.., qn') are the endpoint of 12 and the point A2 e 12 is (ql, on, on'..., q.')- So the point (ql ,--., qi', qi+{,..-, q.') and the point (ql ,..-, qi÷, qi+{,..., q.') are the endpoints of ii and the point _ e Ii is (ql ,.-., qi, qi+{,.--, q.) and so forth. Algorithm: i. Map any edge, 1I, of Q C Rn to the line segment i,' in C. , If the line through 0 and A t' is not perpendicular to Il' then we are finished. o Determine the point Al' on 1I' that is closest to O, if the i line through 0 and Al' is perpendicular to 1I. a) Once (q{, on, ..., q.) and (q,+, on, -.-, qn-) is mapped to C, say f(ql±)=f(ql ±, on', ..., qn') and g (ql±)=g (ql+, on-, ..., q.) b) Then the slope rn = g-_.) - o(a_i f(q{) - f(ql ÷) is calculated. (NASA-CR- 194131) STABILITY BY N94-1086Z LINEAR PROCESSES (Texas Univ.) 9 p Unclas G3/61 0182908 c) The closest point A,' (x,y) to O in C is x- mf_q,) - q(qlll (m + l/m) y = --x/m d) Consider the preimage Al of At'. It is easy to determine the point Al lies on the edge iI on Q by linearity. • Construct the line 12 ± 11 through AI in Q along with ith axis. a) Since all of the q's are fixed except qi then the ith coordinate of A, is founded by qi = x - all linear factors not containinq ___ all linear factor containing qi . Repeat the process for the lines li, i = 3,...,n as was done for 12. The speed of the algorithm can be determined by examining the best/worst case scenario. The best case is that the first edge tried does not have a perpendicular through O which means we exit the algorithm. The worst case is that all of the n-i directional edges have perpendiculars. This requires n iterations. Therefore the average case requires (n+l)/2 iterations• One iteration includes the following calculations: i. f and g for an edge 2. the slope of the line segment in C 3. the point (x,y) which is perpendicular to the line through O 4. determining if (x,y) lies within the line segment 5. mapping (x,y) to the preimage Conjecture I: P(j_,q) _ Hurwitz ¥ q _ Q Iff for some i e {l,...,n} the line through O is not perpendicular to ii' Conjecture 2: p(j_,q) _ Hurwitz V q E Q Iff di = II "- oli, l,...,n then di > di+, for each i = l,...,n-l. To prove the above conjectures we are considering the following direction. Lemma i: If for all edges of Q map to line segment in C no perpendicular to these segments pass through O, then all line segments ii', i=2,...,n do not have a perpendicular through O. Lemma 2: Given any line segment lj in Q parallel jth axis consider all line segments lj÷1_ , k=l,2,...,n-i such that lj+lk along with (j+l) axis then lj÷1k' are all on the same side of lj'. Lemma 1 has already been proven. Example : f(ql,q2,q3,q4) = ql q2 q3 _ + qt + q3 + q2 q3 + 1 g(ql,qe,q3,q4) = ql q2 q3 + 2 qt + q2 + q3 + q4 ist pass endpoints (-2,-2,-2,-2) (-2,-2,-2, 2) image in C (17,-18) (-15,-14) ± point (-1.953846,-15.63077) inside line segment 2nd pass (-2,-2,-2,.3692307) (2,-2,-2,.3692307) (-1.953846,-15.63077) (7.953846,8.36923) ± point (3.843794,-1.586797) inside line segment 3rd pass (.340662,-2,-2,.3692307) (.340662,-2,2,.3692307) (3.843794,-1.586797) (-1.16247,-.3120933) ± point inside (-.1454044 -.5710602 ) inside line segment 4th pass (.340662,-2,1.187366,.3692307) (.340662,2,1.187366,.3692307) (-.1454044,-.5710603) (5.20146,5.046901) ± point inside (.2088863,-.1988064) inside line segment Then O _ Im{p(s,Q) } ,r.,N c_ _J .r-I 0 l I / l / / /J r l t J l f(ql,q2,q3,q_) = qlq2q3q4 + ql + q2 + q3 + q2q3 - 2qlq2q3 - i0 g(qt,q2,q3,q4) = -qlq2q3 + 2ql + q2 + q3 + q_ - qeq3 - q3q4 + 2 ist pass endpoints (-2,-2,-2,-2) (-2,-2,-2,2) image in C (20,-8) (-12,4) ± point (-.1643836,-.4383562) inside line segment 2nd pass (-2,-2,-2,.5205479) (2,-2,-2,.5205479) (-.1643834,-.4383562) (-19.83562,-8.438356) ± point (.1296431,-.31878) outside line segment Then O _ Im{p(s,Q)} IiiII _4 "-.r." ! i f i ] $ ,.O qq r-_ l l / 0 1 d ,I i i i I # i I I