偏微分方程数值解期末试题及答案(2)

?i2?21?a(?i,?j)??2?2,i?j

?i?j?0,而(x?x,?j)??x(x?1)sin(j?x)dx?0212(j?)3[(?1)j?1]

于是得到

最后得到

(x2?x,??cj)??8a(?j?)3(1?j2?2)j为奇数j???(j,?j)??0j为偶数[n?1u2]n(x)?x???8sin[(2k?1)?x]332 (4k?1(2k?1)?[1?(2k?1)]

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