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By Lerner M. E., Repin O. A.

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Additional info for A boundary value problem for mixed-type equations in domains with multiply connected hyperbolicity subdomains

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A. nichtlinearen) Gleichungssystems f (t1 ; x) = s1 .. 36) f (tm ; x) = sm den n dimensionalen L¨osungsvektor zu bestimmen scheitern muß. B. A. nicht l¨osbares) LGS Cx = s , mit C ∈ IRm×n , m ≥ n und s ∈ IRm . 42. 35) erhalten wir etwa  1 t1 t21  . ..   . C= .   . 1 tm t2m   s1  .   s=  ..  . sm  und Anstelle also den (vermutlich vergeblichen Versuch) zu unternehmen, eine exakte L¨osung x des Systems Cx = s zu finden, begn¨ ugen wir uns damit, ein x zu finden, so daß Cx ‘m¨oglichst nahe’ bei s liegt.

A= . . .      ∗ ∗ ∗ ∗ ∗      ∗ ∗ ∗ ∗   ∗ ∗ ∗ hat die Bandbreite m = 2. 25. Besetzt die Matrix A mit der Bandbreite m eine LR-Zerlegung A = LR, so haben L und R ebenfalls die Bandbreite m, denn es gilt lik = 0 f¨ ur i, k mit i − k > m, rik = 0 f¨ ur i, k mit k − i > m. 26. Die Linksdreiecksmatrix L der Cholesky-Zerlegung A = LLT einer positiv definiten Bandmatrix mit der Bandbreite m besitzt ebenfalls Bandbreite m. 50 Lineare Gleichungssysteme Besonders einfach zu behandelnde LGS erh¨alt man f¨ ur tridiagonale Matrizen A der Bandbreite m = 1.

Das LGS Ax = b Empfindlichkeit der L¨osung x gegen¨ heißt schlecht konditioniert, wenn cond(A) >> 1. 34. 01 0 1 0 , , ∆b = b= 0 0 1 1 , , x= x + ∆x = 1 0 200/101 −100/101 54 Lineare Gleichungssysteme Obwohl der Fehler in A bei 1% liegt, haben x, x + ∆x nichts mehr miteinander zu tun. Erkl¨arung: −99 A−1 = ||A||∞ = 2, 100 ||A−1 ||∞ = 200, , 100 −100 cond∞ (A) = 400! Geometrisch: Die Zeilenvektoren a1 , a2 von A haben beinahe die gleiche Richtung. 2 Skalierung Die Kondition eines Problems kann ggf.

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