Tham khảo tài liệu 'numerical methods for ordinary dierential equations episode 13', kỹ thuật - công nghệ, cơ khí - chế tạo máy phục vụ nhu cầu học tập, nghiên cứu và làm việc hiệu quả | 404 NUMERICAL METHODS FOR ORDINARY DIFFERENTIAL EQUATIONS Figure 522 ii Homotopy from an order 3 to an order 4 approximation b pole on left c pole on right a up arrow vertical Figure 522 iii Illustrating the impossibility of A-stable methods with 2v0 p 2 avoid overlapping lines. For t 0 a new arrow is introduced this is shown as a prominent line. As t approaches 1 it moves into position as an additional up arrow to 0 and an additional up arrow away from 0. In such a homotopic sequence as this it is not possible that an up arrow associated with a pole is detached from 0 because either this would mean a loss of order or else the new arrow would have to pass through 0 to compensate for this. However at the instant when this happens the order would have been raised to p which is impossible because of the uniqueness of the v0 V1 . vk approximation. To complete this outline proof we recall the identical final step in the proof of Theorem 355G which is illustrated in Figure 522 iii . If 2vo p 2 then the up arrows which terminate at poles subtend an angle v0 1 2n p 1 n. If this angle is n as in a in this figure then there will be an up arrow leaving 0 in a direction tangential to the imaginary axis. Thus there will be points on the imaginary axis where w 1. In the case of b an up arrow terminates at a pole in the left half-plane again making A-stability impossible. Finally in c where an up arrow leaves 0 and passes into the left half-plane but returns to the right half-plane to terminate at a pole it must have crossed the imaginary axis. Hence as in a there are points on the imaginary axis where w 1 and A-stability is not possible. GENERAL LINEAR METHODS 405 523 Non-linear stability We will consider an example of an A-stable linear multistep method based on the function 1 z w2 2 4 z w 2 4 z . As a linear multistep method this is dH 4 j 1 0 1 0 1 2 1 0 0 1 2 0 1 4 0 0 0 3 4 0 0 0 0 0 where the input to step n consists of the vectors yn i yn 2 hf yn 1 hf yn 2 respectively. .
