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By E. Gekeler

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Extra resources for Discretization Methods for Stable Initial Value Problems

Example text

Proof. See Widlund [67 ]. 4. In o r d e r t o deduce necessary and s u f f i c i e n t the Routh-Hurwitz c r i t e r i o n . 4) Sk- I Sk- 3 . . sk Sk_ 2 . . Ap = . . . . Sk- 3 . . Sk_ 2 . . 0 0 Sk_ I . . I . e . I . I we now ] -k+1 ' S-k+ 2 Sk- I , for Ao-stability s sk , conditions Let Ap denote the Hurwitz matrix o f the p o l y n o m i a l . I , m = 2[k/2], s i = 0 f o r i < O, . 0 oJ U then this well-known result reads as follows; cf. g. 5) Lemma. ,O. implies k i Lemma. ,k. Both r e s u l t s cannot be a p p l i e d d i r e c t l y t o the p o l y n o m i a l ~ ( ¢ , n ) therefore the M6bius transformation i s g e n e r a l l y i n t r o d u c e d in t h i s c o n t e x t , z = (¢ + I)I(~ - I), ¢ = (z + 1)/(z - I), which maps the u n i t d i s k o f the c - p l a n e onto the l e f t half-plane.

Moreover, ~(O,n*) = a 0 - n*b 0 m 0 since otherwise p and s in ~(¢,n) = o(~) - no(t) would have a common f a c t o r . Finally, l e t z be a r o o t of ~(z,n*) with Rez~ 0 and Imz > O. Then 7 is a r o o t , too. But then ¢ = (z + 1 ) / ( z - I) and ~ are two roots of ~(¢,n*) of the same modulus. Hence n* does not belong to the region of r e l a t i v e s t a b i l i t y sequently, the roots of ~(z,n*) l i e in the l e f t which is a c o n t r a d i c t i o n . Con- h a l f - p l a n e , Rez < 0 and the roots of ~(¢,n*) are less than one in absolute value f o r n* c (-b, 0).

41 then ~i(n 2) = ~k_i(q2), i = 0 . . . k, and ~(C,n 2) = ~ k ~ ( c - l , n 2 ) . 4) with a p e r i o d i c i t y i n t e r v a l is necessarily symmetric. Moreover, i f the p r i n c i p a l root ~1,2(0) = I is the only double root of ~(~,0) = pO(~) then t h i s necessary condition is also s u f f i c i e n t . 25) essentially f o r l i n e a r symmetric methods. 28) Lemma. ) Let the linear multistep method defined by ~(~,q2) be convergent and syn~netric, Then it has a periodicity interval iffall growth parameters ^.

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